







X 



m 

■ 



r»»* 










■ 

■ 






iLi 



I 



•«\ 



i> 







MACHINISTS' 



AND 

DRAFTSMEN'S 
HANDBOOK: 

CONTAINING TABLES, RULES AND FORMULAS, 

WITH 

NUMEROUS EXAMPLES EXPLAINING THE PRINCIPLES OF MATHEMATICS 
AND MECHANICS AS APPLIED TO THE MECHANICAL TRADES 

INTENDED AS A 

REFERENCE BOOK FOR ALL 

INTERESTED IN MECHANICAL WORK. 




PEDER ; LOBBEN, 

MECHANICAL ENGINEER. 

Member of the American Society of Mechanical Engineers and Worcester 

County Mechanics Association. Honorary Member of the N. A. 

Stationary Engineers. Non-Resident Member of the 

Franklin Institute. 



NEW YORK: 
D. VAN NOSTRAND COMPANY, 

23 Murray and 27 Warren Streets. 
1900. 



TWO COPIES RECEIVED. 



Library of Cdisgeifty 
Qfftaa of $fca 

APR ] 3 1900 

Kegl«t«r of Copyrfg&f^ 



56628 

Copyright, 1899, 

BY 

PEDER LOBBEN. 



^ 



SECOND COPY, 






PREFACE. 

It is the author's hope and desire that this book, which is 
the outcome of years of study, work and observation, may be 
a help to the class of people to which he himself has the honor 
to belong, — the working mechanics of the world. 

This is not intended solely as a reference book, but it may 
also be studied advantageously by the ambitious young engineer 
and machinist; and, therefore, as far as believed practical 
within the scope of the work, the fundamental principles upon 
which the rules and formulas rest are given and explained. 

The use of abstruse theories and complicated formulas is 
avoided, as it is thought preferable to sacrifice scientific hair- 
splitting and be satisfied with rules and formulas which will 
give intelligent approximations within practical limits, rather 
than to go into intricate and complicated formulas which can 
hardly be handled except by mathematical and mechanical 
experts. 

In practical work everyone knows it is far more important 
to understand the correct principles and requirements of the job 
in hand than to be able to make elaborate scientific demonstra- 
tions of the subject; in short, it is only results which count in 
the commercial world, and every young mechanic must remem- 
ber that few employers will pay for science only. What they 
want is practical science. Should, therefore, scientific men, (for 
whom the author has the greatest respect, as it is to the scien- 
tific investigators that the working mechanics are indebted for 
their progress in utilizing the forces of nature), — find nothing of 
interest in the book, they will kindly remember that the author 
does not pretend it to be of scientific interest, and they will 
therefore, in criticizing both the book and the author, remember 
that the work was not written with the desire to show the reader 
how vulgarly or how scientifically he could handle the subject, 
but with the sole desire to promote and assist the ambitious 
young working mechanic in the world's march of progress. 

P. LOBBEN. 

New York, October, 1899. 



IRotes on flfeatbematics. 



A Unit is any quantity represented by a single thing, as a 
magnitude, or a number regarded as one undivided whole. 

Numbers are the measure of the relation between quanti- 
ties of things of the same kind and are expressed by figures. 

Numbers which are capable of being divided by two without 
a remainder are called even numbers. 2, 4, 6, 8, etc., are 
even numbers. 

Numbers which are not capable of division by two without 
giving a remainder are called odd numbers. 1, 3, 5, 7, 9, etc., 
are odd numbers. 

A number which can not be divided by any whole number 
but itself and the number 1 without giving a remainder is 
called a prime number. 1, 2, 3, 5, 7, 11, 13, 17, 19, etc , are 
prime numbers. 

All numbers that are not prime are said to be composite 
numbers, because they are composed of two or more factors ; 
4, 6, 8, 9, 10, 12, etc., are composite numbers. 

Whole numbers are called integers. Whole numbers are 
also called integral numbers. 

A mixed number is the sum of a whole number and a 
fraction. 

The least common multiple of several given numbers is 
the smallest number that can be divided by each without a 
remainder. For instance, the least common multiple of 3, 4, 6, 
and 5 is 60, because 60 is the smallest number that can be 
divided by those numbers without a remainder. 

Signs. 

+ (plus) is the sign of addition. 

— (minus or less) is the sign of subtraction. 

The signs + and — are also used to indicate positive and 
negative quantities. 

X (times or multiply) is the sign of multiplication, but in- 
stead of this sign, sometimes a single point (.) is used, especially 
in formulas; in algebraic expressions very frequently factors 
are written without any signs at all between them. For in- 
stance, aXb or a.b or ab. All these three expressions indicate 
that the quantity # is to be multiplied by the quantity b. 



2 NOTES ON MATHEMATICS. 

-r- (divided by) is the sign of division. 

== (equal). When this sign is placed between two quanti- 
ties, it indicates that they are of equal value. For instance: 

4 + 5 + 2 = 11 
8—3+6—2=9 
8 X 12 = 96 

100 -r- 5 = 20 

. (decimal point) signifies that the number written after it 
has some power of 10 for its denominator. 

° ' " means degrees, minutes and seconds of an angle. 
' " means feet and inches. 

a' a 1 ' a'" reads a prime, a second, a third. 

a\ a% as reads a sub 1, a sub 2, a sub 3, and is always 
used to designate corresponding values of the same element. 

n 

V This is the radical sign and signifies that a root is to be 
extracted of the quantity coming under the sign ; this maybe 
square root, cube root, or any other root, according to what 
there is signified by the number prefixed in place of the letter n . 

3 4 

For instance : V reads square root, VVeads cube root, V reads 

5 

fourth root, V reads fifth root, V64 = S, because 8 X 8 = 64 

3 

*/64 = 4, because 4 X 4 X 4 = 64 

\/sT = 3, because 3X3X3X3 = 81 
The sign that a quantity is to be raised to a certain 
power is a small number placed at the upper right hand corner 
of the quantity ; this number is called the exponent. For in- 
stance, 7* 2 signifies that 7 is to be squared or multiplied by itself, 
that is : 

T 2 = 7 X 7 = 49 
73 = 7x7X7 = 343, etc. 

< [ braces, [ ] brackets, ( ) parentheses, signify that 

the quantities which they include are to be considered as one 
quantity. For instance : 35 — (8 + 6) is equal to 35 — 14 = 21. 
In this case the parenthesis indicates that not only 8, but the 
sum of 8 + 6 is to be subtracted from 35. 

(vinculum or bar) is a straight line placed over 

two or more quantities, indicating that they are to be operated 
upon as one quantity. For instance, V25 + 11. The vinculum 
attached to the radical sign indicates that the square root shall 
be extracted from the sum of 25+11, which is the same as the 
square root of 36. 

35 I 15 1 22 
In an expression as . — X — J_ — the bar indicates that 

F 3X8 



NOTES ON MATHEMATICS. 3 

the sum of 35+15+22 shall be divided by the product of 3XS 
which is the same as 72 divided by 24. 

Whenever a number or a quantity is placed over a line 
and a number or a quantity is placed under the same line it 
always indicates that the number or quantity over the line shall 
be divided by the number or quantity under the line. Such a 
quantity is called a fraction. 

The quantity above the line is called the numerator, and 
the quantity below the line is called the denominator. A frac- 
tion may be either proper or improper. The fraction is proper 
when the numerator is smaller than the denominator; for 
instance, |; but improper if the numerator is larger than the 
denominator, for instance, V — If- 

A fraction can always be considered simply as a problem 
in division. 

Formulas. 

A formula is an algebraic expression for some general 
rule, law or principle. Formulas are used in mechanical 
books, because they are much more convenient than rules. 
Generally speaking, the knowledge of algebra is not required 
for the use of formulas, because the numerical values corre- 
sponding to the conditions of the problem are inserted for 
every letter in the formula except the letter representing the 
unknown quantity, which then is obtained by simple arithmet- 
ical calculations. It is generally most convenient to begin the 
interpretation of formulas from the right-hand side; for instance, 
the formula for the velocity of water in long pipes is : 



= 8.02^; 



ry — h. CI. 



In this formula v represents the velocity of the water in 
feet per second. 

h represents the "head"* in feet. 

cl represents the diameter of the pipe in feet. 

f represents the friction factor determined by experiments. 

/ represents the length of the pipe in feet, and 8.02 is a 
constant equal to the square root ot twice acceleration due to 
gravity. 

Assume, for instance, that it is required to find the velocity 
of the flow of water in a pipe of 3 inches diameter ( % foot) ; 
the length of the pipe is 1,440 feet, the "head" is 9 feet, and 
the friction factor is 0.025. 

Inserting in the formula these numerical values, and 
for convenience writing the diameter of the pipe in decimals, 
we have : 



* In hydraulics the word " head " means the vertical difference between the 
level of the water at the receiving end of the pipe and the point of discharge, or 
its equivalent in pressure. See Hydraulics, page 413. 



NOTES ON MATHEMATICS. 



V: 



v- 8.02 XJ 9 -X 0.25 



.025 X 1440 
Solving the problem step by step we have : 

v = 8.02 X A /_ 2 - 2 JL 
* 36 

v — 8.02 X \Z0.0625" 

v = 8.02 X 0.25. 

v = 2.005 feet per second. 

In mechanical formulas, if not otherwise specified, it 
is always safe to assume the letter g to mean acceleration 
due to gravity, usually taken as 32.2 feet or 9.82 meters. In 
formulas relating to heat the letter J usually signifies the 
mechanical equivalent of heat = 778 foot pounds of energy; 
but in formulas relating to strength of materials the letter J 
usually signifies the polar moment of inertia, and the letter / 
the least rectangular moment of inertia. The letter x always 
expresses the unknown quantity. The following Greek letters 
are also used more or less. The letter n-, called pi, is used to 
signify the ratio of the circumference to the diameter of a 
circle, and is usually taken as 3.1416. % called sigma, usually 
signifies the sum of a number of quantities. The letter A, 
called delta, usually signifies small increments of matter. 

The letter 0, called theta, or the lefter 3>, called phi, usually 
signifies some particular angle, sometimes also the coefficient 
of friction. But all these letters may be employed to express 
anything, although it is usually safe, if not otherwise specified, 
to expect their meaning to be as stated. It is always customary 
to express known quantities by the first letters in the alphabet, 
such as a, fr, c, etc., and unknown quantities by such letters as 
x,y, z, etc. 



Hritbmetic. 



Addition. 



All quantities to be added must be of the same unit; we can 
not add 3 feet + 8 inches + 2 meters, without first reducing 
these three terms either to feet, inches or meters. The same 
also with numbers. Units must be added to units, tens to tens, 
hundreds to hundreds, etc. 

Example. 

318 + 5 + 38 + 10 + 115 = 4S6 

Solution : 318 

5 

38 

10 

115 



486 = Sum. 

Subtraction. 

Two quantities to be subtracted must be of the same 
unit. 

In subtraction, the same as in addition, the units are 
placed under each other, and units are subtracted from units, 
tens from tens, hundreds from hundreds, etc. 

Example. 

2543 — 1828 = 715 
Solution: 2543 . . Minuend. 

1828 . . Subtrahend. 



715 . . Difference. 
Subtrahend + Difference = Minuend. 
(5) 



O ARITHMETIC. 

riultiplication. 

A quantity is multiplied by a number by adding it to itself 
as many times as the number indicates. 

Example. 

314 X 3 = 314 + 314 -f- 314 = 942 

Solution: 314 . . Multiplicand)^ . 

3 . . Multiplier [Factors. 

942 . . Product. 

Product 
Multiplicand = MultipEer. 

Product 
Multiplier = Multiplicand. 



Division. 

The quantity or number to be divided is called the dividend. 
The number by which we divide is called the divisor. The 
number that shows how many times the divisor is contained in 
the dividend is called the quotient. 

Example. 

6852 — 3 = 2284 

Solution: 3)6852(2284 6852 . . Dividend. 

6 3 . . Divisor. 

— 2284 . . Quotient. 

8 Divisor X Quotient = Dividend. 



25 
24 

12 

12 



ARITHMETIC. 



FRACTIONS. 



Addition. 

Fractions to be added must have a common denominator ; 
thus we cannot add y 2 -f- % + X ~r~ H unless they be reduced 
to a common denominator instead of the denominators two, 
three and four ; in other words, we must find the least common 
multiple of the numbers 2, 3 and 4, which is 12. Thus we 
have : 




26 _ 9-2 9i 

12 — *1 2 ^6 



Example 2. 



Add: T V + I + i + tV + I + f 4- I + 



The common denominator is found in the following 
manner: Write in a line all the denominators, and divide with 
the prime number, 2, as many numbers as can be divided with- 
out a remainder. The numbers that cannot be divided without 
a remainder remain unchanged, and these together with the 
quotients of the divided numbers, are written in the next line 
below. Repeat this operation as long as more than one num- 
ber can be divided without remainder, then try to divide 
by the next prime number, and so on. These divisors and all 
those numbers remaining undivided in the last line are multi- 
plied together, and the product is the least common denominator. 

2) n % i w % 7 ? 9 



2) 


? i 2 


3 7 4" 9 


2) 


i 2 1 


3 3 7 2 9 


3) 


2 11 


3 3 7 1 jf 



211 11713 

The common denominator is thus: 

2X2X2X3X2-X7X3= 1008 

Thus 1008 is the least common multiple of 16, 8, 4, 12, 7 
and 9. 



8 



ARITHMETIC. 



The principle of this solution can probably be better 
understood by resolving these numbers into prime numbers, 
and also resolving 1008 into prime numbers ; we then find that 
1008 contains all the prime numbers necessary to make 16, 8, 4, 
12, 6, 7 and 9. 

Prime numbers in 1008 are 2 2 2 2 7 3 



n 1008 


are 


2 


2 


2 


2 


" 16 


u 


2 


2 


2 


2 


8 


a 


2 


2 


2 




4 


a 


2 


2 






« 12 


a 


2 


2 


3 




6 


u 


2 


3 






7 


a 


7 








" 9 


u 


3 


3 







Solution of Example 2 

1008 



r 7 e 63 X 7 = 441 
| 126 X 5 = 630 
252 X 1 = 252 

84 X 7 = 588 
168 X 5 = 840 
144 X 5 = 720 
126 X 3 = 378 
112 X 4 == 448 



l 

4 

7 

T2 

5 

e 

5 
7 
.3 
8 
4 
9 



2_6_5_ 



42U< = TOO? 4x008 



Subtraction. 



When fractions are to be subtracted, they must first be 
reduced to a common denominator, the same as in addition. 

Example. 

|- — J. must be reduced to f — f = f 
Examples. 



No. 


1. 


5 

8 


1 
4 


No. 


2. 


TJ 


1 

5 


No. 


3. 


1 1 
1 6 


5 
3 



1 2 




23. 

60 

- 11 
_ 32 



FRACTIONS. 



Multiplication. 



Fractions are multiplied by fractions, by multiplying numer- 
ator by numerator and denominator by denominator ; thus : 



8 ^ 12 i) 6 3 3~ 

The correctness of this rule can easily be understood if 
we consider these two fractions as two problems in division, 
f X T v will then be 3 divided by 8 and the quotient multiplied 
by 7 and the product divided by 12; thus, 3 is to be multiplied 
by 7 and the product is to be divided by S times 12. Therefore : 

8 *- 3 "8X/2 8X 4 3 ' 2 

A mixed number may first be reduced to an improper frac- 
tion and then multiplied as a common fraction, numerator by 
numerator and denominator by denominator. For instance : 

ql v :; — . I v- 3 — 2J, — 9 5 
O, A T 2*4 S Z 8 

A fraction may be multiplied by a whole number by multi- 
plying the numerator and letting the denominator remain un- 
changed. For instance : 

- 7 - x -2 — i± — 1 - 2 - — 11 

This must be correct, because we may consider 7 as indicat- 
ing the quantity and 12 as indicating what kind of quantity in 
exactly the same sense as we may say 7 dollars or 7 cents : if 
either of those were multiplied by 2 the product would, of 
course, be either dollars or cents respectively, and for the same 
reason 7 twelfths multiplied by 2 must be 14 twelfths. 

A fraction may also be multiplied by a whole number, by 
dividing the denominator by the number and letting the numer- 
ator remain unchanged. For instance : 

T 7 2- X 2 = |- = 1}, because T 2 2 is equal to £, so must T V X 2 

6 - L 6 

Examples. 

No. 1. 3iXf=^X! = ff 

No. 2. H X H= f X | = f = 2# 

No. 3. A X | = ^ 

No. 4. If X A = ¥ X A = ¥ X T V = { X i = 1 



IO FRACTIONS. 

Division. 

A fraction is divided by a fraction by writing the fractions 
after each other, then inverting the divisor (that is, changing 
its numerator to denominator and its denominator to numer- 
ator), proceed as in multiplication. For instance : 

5. -i_ 3. 5. V 4. 2.0. 5 

8 • 4 8^3 24 6 

Tlie reason for this rule can very easily be understood, 
when we consider the fractions as problems in division. That is 
to say, 5 shall be divided by 8 and the quotient is to be divided 
by one-fourth of 3. But if the quantity f is divided by 3 instead 
of one-fourth of 3, we must, of course, multiply the quotient by 
4 to make the result correct. Therefore : 

5. v 4 

5- -L. 3. 8 ^ _2_0 • O 2.0 5. 

8 • 4 o 8 ■ ° 24 6 

A fraction may be divided by a whole number by dividing 
the numerator by the number and letting the denominator re- 
main unchanged. For instance : 

_9_ _i_ -.-> _3_ 

16 • ° 16 

A fraction may be divided by a whole number by multiply- 
ing the denominator by the whole number and letting the 
numerator remain unchanged. For instance : 

2 _i_ q — 2 

3 • ° 9 

Mixed numbers are reduced to improper fractions the 
same as in multiplication ; they are then figured the same as if 
they were proper fractions. 



1 — T_V2 — 14 — 1 

2 16 <*■ 1 16 8 

3 _5_ V A 2.0 10 

4 18 ^ 3 54 27 

• 1 1 17 V 3. 5.1 

• x 3_ 8 ^ .4 32 

-f- 6 = | -^ 6 = xV 

In No. 4 it will be understood that | divided by 6 must be 
T %, because x 1 ^ is exactly a sixth of ^. 

In No. 5, also, it will be understood that if V~ is divided by 
4, the quotient must be f, because 4 is one-fourth of 16. 



Examples. 




No. 1. 


16 


No. 2. 


5 

18 


No. 3. 


^8 


No. 4. 


n 


No. 5. 


H 



DECIMALS. I I 



To Reduce a Fraction of One Denomination to a Fraction 

of Another Fixed Denomination, and Approx= 

imately of the Same Value. 

In mechanical calculations, on drawings, and on other oc- 
casions, it is very frequently necessary to reduce fractions of 
other denominations to eighths, sixteenths, thirty-seconds T or 
sixty-fourths. This may be done by multiplying the numerator 
and the denominator of the given fraction by the number which 
is to be the denominator in the new fraction, then dividing this 
new numerator and denominator by the denominator of the 
given fraction. 

Example. 

Reduce % to eighths, sixteenths, thirty-seconds, sixty- 
fourths, or to hundredths. 

51/ 
I X f = M — o or H approximately. 

10? 
I X t! = ft = -Tjf^ or ii approximately. 



16 

]2 



214 
X |f = ff = ^f- or |4 approximately. 

42- 2 - 
f X f| = f|| = -~ or H approximately. 

PC* 2 

I X iU = lt§ = Jq ~ or toV approximately. 

Thus f, instead of f , is considerably too small, namely, jL 
but || is a great deal nearer, only T ^2 too large, and 0.67 is 3^ 
too large. 



DECIMALS. 

In decimal fractions the denominator is always some power 
of ten, such as tenths, hundredths, thousandths, etc. 

The denominator is never written, as it is fixed by the rule 
that it is 1 with as many ciphers annexed as there are figures 
on the right-hand side of the decimal point. 

y z = 0.5 = five-tenths = T \ 

% = 0.25 = twenty-five hundredths = -ffa 

yi = 0.125 = one hundred and twenty-five thousandths = T Wo 
\ l / 2 = 1.5 = one and five-tenths = 1 T \ 
1%. — 1.25 == one and twenty-five hundredths = l T Vo> etc. 



: : DECIMAL 

Figures on the left side of the decimal point are whole num- 
bers. When there are no whole numbers, sometimes a cipher 
is written on the left side of the decimal point but this is not 
always done, as it is common with many writers not to write 
anything on the left side of the decimal point when there is 
no whole number. 

Thus : 

: : may be written .5 
: _ ..' .. .. _- 

H -125 

It is. however, preferable to nil in a cipher on the left- 
hand side of the decimal point when there is no whole number, 
as by so doing the mistake of reading a decimal for a whole 
number is prevented. 

To Reduce a Vulgar Fraction to a Decimal Fraction. 

Annex a sufficient number of ciphers to the numerator, 
divide the numerator by the denominator, and point off as many 
decimals in the quotient as there are ciphers annexed to the 
numerator. 

Example. 

Reduce '/% to a decimal fraction. 

Solution : 

- 7.00c i mi 

64 






40 
40 



00 

Thus, '/% is equal to the decimal fraction . -"" 



DECIMALS. 



13 





Fractions Reduced to Exact Decimals. 




1 

6 4 


.015625 


1 7 

6~4 


.265625 


33 

6~4 


.515625 


49 

6 4 


.765625 


1 

32 
3 
6 4 


.03125 

.046875 


9 

32 
1 9 
6~4 


.28125 
.296875 


1 7 
32 
3 5 
6 4 


.53125 

.546875 


2 5 

3 2 
5 1 
6~4 


.78125 
.796875 


1 


.0625 


5 

1 6 


.3125 


9 
1 6 


.5625 


1 3 
T6" 

5 3 

6 4 


.8125 

.828125 


5 
6 4 


.078125 


2 1 

64 


.328125 


3 7 
6~4 


.578125 


3 
3 2 

7 
6 4 


.09375 
.109375 


1 1 
32 
23 
6 4 


.34375 
.359375 


19 
3 2 
39 
6 4 


.59375 
.609375 


2 7 

3 2 
55 
6 4 


.84375 
.859375 


1 

8 


.125 


3 

8 


.375 


5 

8 


.625 


7 
"8" 


.875 


9 
6 4 


.140625 


2 5 
6 4 


.390625 


41 

6 4 


.640625 


5 7 

6 4 


.890625 


5 

3 2 


.15625 


13 
3 2 


.40625 


2 1 

3 2 


.65625 


29 
3 2 


.90625 


1 1 

64 

3 

T6~ 

13 

6~4 


.171875 
.1875 

.203125 


2 7 

64 

7 

16 


.421875 
.4375 


43 
64 
1 1 
16 


.671875 

.6875 


59 
"6~4 
1 5 
1 6 

6 1 

6~4 


.921875 
.9375 

.953125 


29 

6 4 


.453125 


45 

6 4 


.703125 


7 

32 
1 5 

6 4 


.21875 
.234375 


1 5 
3 2 
3 1 
6~4 


.46875 
.484375 


2 3 

3 2 
£7 
6 4 


.71875 
.734375 


31 
32 
63 
6 4 


.96875 
.984375 


1 

4 


.25 


1 
2 


.5 


3 
4 


.75 


1 


1. 



To Reduce a Decimal Fraction to a Vulgar Fraction. 

Write the decimal as the numerator of the fraction and set 
under it for the denominator the figure one, followed by as 
many ciphers as there are decimal places ; then cancel the frac- 
tion thus written, to its smallest possible terms. 

Example. 

Reduce 0.3125 to a vulgar fraction. 

Solution : 

0.3125 = T 3 oWo> cancelling this by five we have ^Wo = iM 

2 5 3_ 

80 16' 



To Reduce a Decimal Fraction to a Given Vulgar Fraction 
of Approximately the Same Value. 

Multiply the decimal by the number which is denomi- 
nator in the fraction to which the decimal shall be reduced, 
and the product is the numerator in the fraction. 

Example. 

Reduce 0.48437 to sixteenths, thirty-seconds and sixty- 
fourths. 



14 DECIMALS. 

Solution : 

0.484375 X 16 = 7.75, gives 7^_, or T ?_, approximately. 
0.484375 X 32 = 15.5, gives J*^, or 15, approximately. 
0.484375 X 64 = 31, gives si exactly. 

If the result does not need to be very exact, probably T %, 
which is g 3 T too small, is near enough, or the result, - 7 jl r \ may be 
called }4, which is g x T too large, £§ is g 1 ^ too small, therefore 
either ]/ 2 or \\ is only g : T different from the true value. The 
first is g 1 ^ too large and the last is -^ T too small, and which 
fraction, if either, should be preferred, will depend entirely upon 
the purpose for which the problem is solved. §£ is the exact 
value. 

Addition of Decimal Fractions. 

In adding decimal fractions, care should be taken to place 
the decimal points under each other ; then add as if they were 
whole numbers. 

Example. 

Add 50.5 + 5.05 + 0.505 + 0.0505 
Solution : 

50.5 
5.05 
0.505 
0.0505 

56.1055 

To prevent mistakes and mixing up of the figures during 
addition, it is preferable to make all the decimal fractions in 
the problem of the same denomination by annexing ciphers. 

Thus : 50.5000 

5.0500 
0.5050 
0.0505 

56.1055 

Subtraction of Decimal Fractions. 

The decimal point in the subtrahend must be placed under 
that in the minuend ; the fractions are both brought to the 
same denomination by annexing ciphers, then the subtraction 
is performed just as if they were whole numbers, but close 
attention must be paid to have the decimal point in the same 
place in the difference as it is in the minuend and subtrahend. 

Example. 

318.05 — 121.6542 



DECIMALS. 15 

Solution : 318.0500 Minuend. 

121.6542 Subtrahend. 
190.3958 Difference . 

riultiplication of Decimal Fractions. 

Multiply the factors as if they were whole numbers. After 
multiplication is performed, count the number of decimals in 
both multiplier and multiplicand and point off (from the right) 
the same number of decimals in the product. 

If there are not enough figures in the product to give as 
many decimals as required, then prefix ciphers on the left until 
the required number of decimals is obtained. 

Example 1. 

0.08 X 0.005 = 0.00520 = 0.0052 

In this example it is necessary after the multiplication is 
performed, to prefix two ciphers to the product in order 
to obtain the necessary number of decimals, because the pro- 
duct, 520, consists of only three figures, but the two numbers, 
0.08 and 0.065, contain five decimals. 

Example No. 2. 

3.1416 X5 = 15.7080 = 15.708 
Example No. 3. 

3.1416 X 0.5 = 1.57080 = 1.5708 

Division of Decimal Fractions. 

Divide same as in whole numbers, and point off in the 
quotient as many decimals as the number of decimals in the 
dividend exceeds the number of decimals in the divisor. 

If the divisor contains more decimals than the dividend, 
then before dividing annex ciphers (on the right-hand side) in 
the dividend until dividend and divisor are both of the same 
denomination, then the quotient will be a whole number. 

Example. 

43.62 -f- 0.003 = 14,540 

Solution : 0.003 ) 43.620 ( 14,540 

3 

13 
12 

16 
15 

12 
12 

00 



1 6 RATIO AND PROPORTION. 

In this example the dividend consists of only two decimals, 
but the divisor has three, therefore we have to annex a cipher 
to the dividend. This brings divisor and dividend to the same 
denomination, and the quotient is a whole number. 

Example 2. 

43.62 -r- 0.3 = 145.4 

In this example the dividend has one decimal more than 
the divisor, therefore the quotient has one decimal. 



RATIO. 

The word ratio causes considerable ambiguity in mechani- 
cal books, as it is frequently used with different meaning by 
different writers. 

The common understanding seems to be that the ratio be- 
tween two quantities is the quotient when the first quantity is 
divided by the last quantity ; for instance, the ratio between 3 
and 12 is %, but the ratio between 12 and 3 is 4. The ratio be- 
tween the circumference of a circle and its diameter is tt or 
3.1416, but the ratio between the diameter and the circumfer- 
ence is \ or 0.3183, etc. This is the sense in which the word is 
used in this book, as this seems to agree with the common cus- 
tom with most mechanical writers. 

The term ratio is also sometimes applied to the difference 
of two quantities as well as to their quotient ; in which case the 
former is called arithmetical ratio, and the latter geometrical 
ratio. (See Progressions, page 68.) 



PROPORTION. 



In simple proportion there are three known quantities by 
which we are able to find the fourth unknown quantity ; there- 
fore proportion is also called "the rule of three", and it is either 
direct or inverse proportion. 

It is called direct proportion if the terms are in such ratio to 
one another that if one is doubled then the other will also have 
to be doubled, or if one is halved the other must also be halved. 
For instance, if 50 pounds of steel cost $25, how much will 250 
pounds cost? 

50 lbs. cost #25; 250 must cost =£ A — = #125. 

50 

This is direct proportion, because the more steel we buy, 
the more money we have to pay. 

In inverse proportion the terms are in such ratio that if one 
is doubled the other is halved, or if one is halved the other is 
doubled. 



proportion. 1 7 

Example. 

Eight men can finish a certain work in 12 days. How many- 
men are required to do the same work in 3 days ? 

Here we see that the fewer days in which the work is to be 
done, the more men are required. Therefore, this example is 
in inverse proportion. 

In 12 days the work was done by 8 men ; therefore, in order 

to do the work in 3 days it will require i_^— _ = 32 men. 

It requires 4 times as many men because the work is to be 
done in one quarter of the time. 

Compound Proportion. 

A proportion is called compound, if to the three terms there 
are combined other terms which must be taken into considera- 
tion in solving the problem. 

A very easy way to solve a compound proportion is to 
(same as is shown in the following examples) place the con- 
ditional proposition under the interrogative sentence, term 
for term, and write x for the unknown quantity in the inter- 
rogative sentence ; draw a vertical line ; place x at the top at the 
left-hand side ; then try term for term and see if they are direct 
or inverse proportionally relative to x, exactly the same way as 
if each term in the conditional proposition and the correspond- 
ing term in the interrogative sentence were terms in a simple 
rule-of-three problem. Arrange each term in the interrogative 
sentence either on the right or left of the vertical line, according 
to whether it is found to be either a multiplier or a divisor, when 
the problem, independent of the other terms, is considered as a 
simple rule-of-three problem. 

After all the terms in the interrogative sentence are thus 
arranged, place each corresponding term in the conditional 
proposition on the opposite side of the vertical line. Then 
clear away all fractions by reducing them to improper frac- 
tions, and let the numerator remain on the same side of the verti- 
cal line where it is, but transfer the denominator to the opposite 
side. Now cancel any term with another on the opposite 
side of the vertical line; then multiply all the quantities on the 
right side of the vertical line with each other. Also multiply 
all the quantities on the left side of the vertical line with each 
other. 

Divide the product on the right side by the product on the 
left, and the quotient is the answer to the problem. 

Example 1. 

A certain work is executed by 15 men in 6 days, by work- 
ing 8 hours each day. How many days would it take to do the 
same amount of work if 12 men are working 7J^ hours each 
day? 



PROPORTION. 



Solution 



15 Men 6 Days 8 Hours. 

12 " x "' ~y 2 " 

x 6 

1 2 12 15 1 

'jy 2 s 



S Days. 



Example 2. 



A steam engine of 25 horse power is using 1500 pounds of 
coal in 1 day of 9)4 working hours. How many pounds of coal 
in the same proportion will be required for 2 steam engines each 
haying 30 horse power, working 6 days of 12% hours each day ? 

Solution : 

1 Machine 25 Hp. 1,500 pounds 1 Day 9}4 hours. 



1 
1 

Example 3. 



30 

19 

l 



X 

1 

23 

1 
9 



x " 6 
1500 300 
2 
30 6 
6 2 

12 2 3 38 2 
o 



12- 



2S.S00 pounds of coal. 



A piece of composition metal which is 12 inches long, 'Z l / 2 
inches thick and \ l / 2 inches wide, weighs 45 pounds. How many 
pounds will another piece of the same alloy weigh, if it meas- 
ures S inches long. 1% inches thick and 6 3 4 inches wide? 

12" long. 3 1 /" thick. 4>£" wide. 45 pounds. - 
S " 1% " 6% " x 





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22)4 pounds. 



INTEREST. 1 9 



INTEREST. 



The money paid for the use of borrowed capital is called 
interest. It is usually figured by the year per 100 of the 
principal. 



Simple Interest. 

Simple interest is computed by multiplying the principal by 
the percentage, by the time, and dividing by 100. 

What is the interest of #125, for 3 years, at 4% per year? 
Solution : 

125 X 4 X 3 



100 



= #15 



In Table No. 1, under the given rate per cent., find the 
interest for the number of years, months, and days ; add these 
together, and multiply by the principal invested, and the 
product is the interest. 

Example. 

What is the interest of #600, invested at 6%, in 5 years, 
3 months, and 6 days ? 

Solution : 

#1.00 in 5 years at 6% = 0.30 
" " 3 months " " = 0.015 
" " 6 days " " = 0.001 

0.316 
600 = Principal. 



#189.60 = Interest 



20 



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2 2 COMPOUND INTEREST. 

Compound Interest Computed Annually. 

If the interest is not withdrawn, but added to the principal, 
so that it will also draw interest it is called compound interest. 

Example. 

What is the amount of $300, in 3 years, at h% ? The 
interest is added to the principal at the end of each year. 

Solution: 

Principal and interest at the end of first }~ear. 

10.") X 300 MiS 

= S31q. 

100 

Principal and interest at the end of second year, 

105 X 315 c .. on _. 
100 

Principal and interest at the end of third year, 

105 X 330.75 



100 



= S347.2875. = S347.29 = Amount. 



When compound interest for a great number of years is to be 
calculated, the above method of figuring will take too much 
time, and the following interest tables. No. 2 and No. 3. are 

computed in order to facilitate such calculations. 

In Table Xo. 2. under the given rate per cent., and opposite 
the given number of years, find the amount of one dollar in- 
vested at that rate for the time taken. Multiply this by the 
principal invested and the product is the amount. 

Example. 

$400 is invested at h% compound interest for 17 years, com- 
puted annually. What is the amount ? 

Solution : 

In Table Xo. 2. under 5%. and opposite 17 years, we find 
2.292011. Multiply this by the principal. 

Thus : 

2.292011 
400 



916.S044 = $916.80 = Amount. 



COMPOUND INTEREST. 



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24 COMPOUND INTEREST. 

Compound Interest Computed Semi=Annually. 

When compound interest is to be computed semi-annually, 
use Table No. 3. Under the given rate and opposite the given 
number of years, find' the amount of one dollar invested and 
interest computed semi-annually for the time taken. Multiply 
this by the principal invested, and the product is the amount. 

Example. 

$350 is put in a savings bank paying 4%, computed semi- 
annually. What is the amount in 10 years ? 

Solution: 

Under 4%, and opposite 10 years, we find the number 
1.4860. This we multiply by the principal invested. 

Thus : 

1.485949 
350 



520.08215 = $520.08 = Amount. 

To compute compound interest for longer time than 
is given in the tables, figure the amount for as long a time 
as the table gives ; then consider this amount as a new princi- 
pal invested, and use the table and figure again for the rest of 
the time. 

Example. 

What is the amount of $40, left in a savings bank 18 years, 
at 4%, and the interest computed semi-annually. The table 
only gives 12 years, therefore we will look opposite 12 years, 
under 4%, and find the number 1.608440. This we multiply by 
the principal invested. 

Thus: 

1.608440 

40 



64.3376 



But now we have to compute for 6 years more, therefore 
under 4%, and opposite 6 years, we find the number 1.268243. 
Multiplying this by the principal, which is now considered as 
being invested 6 years more, we have : 

1.268243 X 64.3376 = $81.60 = Amount. 

Thus, $40, invested at 4% interest, computed semi-annually, 
will, after 18 years of time, amount to $81.60. 



COMPOUND INTEREST. 



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OOrHHHlNtNMMTtiTfiOxJOOt-t-QOOOOH^lN 





Ot-Si*OMOMOMO?l(COiO©H^a(MHTPH 
©(NOt-iOt— ^r-i-iMOSOGGwOTCOOOOSOGOCS 
G I- O (N O CO 1- 1- S n I- iC O O t* Q0"*iOr- !(NH»Ot- 

MO05(Ni:O(NOOii00N!0riK5Ot0OiOOOHl-M 
OOOHHHNINCOMM^^iOiOOOt-t-aOCCOlOO 




i0-JC0O-tOT)(C0OMO(Ni0O00O^WOiOii^O 

CiaHO3 f JDOOCC(10QHt«OO(NOif5rtCCt"HK 

O0fjCfj^OO-HC0OO00i005C0O?0©OC0i0ia?0t- 

OOOMrHCiCOGOOOOWTfOONCO^HOiiXCOOH^CO 

<MiOt-OCO»OQOi-H"*QOi-HT^l^-T-l^QO(MOClCCr-(MOO 

OOOi-iHHHINININMMMit^^OiOrttOCl-t-X 




ococot-iNcowHOr-iinmccioiN^QOOOHfMeocD 

»OOOOOGOiO(X)r^<NCOO-*-^(NCC>t~-Oi-HiOO»OC^r- 

NiCOJCOMNCO^HOt-OOOOl-ONOOiCHOOO 
N^t?0©HT(i©OJ(NTCt-OMOCi(Ni(5 0(NOCiWfflO 




OQO<NHCOaOCOOiOCOO)OOt-OOOt-aoO(NHO 

oococoooocooot-^oan-oo^^^^oixiOTti 

^(N^OrtOOOOlMCNOiiCCt-^INOOO©®©^ 

OHN^oooHinooncoMamfNOixiooiciocco 

OOOOHHHHH(NiM(N(NCOWMi* , *'*^iniO"00 


*5 


0-^0©iM'Mi0^t'001ffiOt-Ot-0'*^(Ct-OH 

O'NiOrtOiNriQ0HCDWQ0H(NiN(NCC'*t-W<N©'* 

iOW*CO©^rt(XiCJTt(!M'*QO:(MOiO»0^ , r-iOt-MTt< 

jt-ioMHOOooocoaoHiN^t-acooo^ffi^o© 

HC0«3t-OOlN^iC00HMiOt-©H^®OrtW©©H 
OOOOOrtHHrHHlMNINNINMMMCO'^Tli'^TJIiO 

rtHrlrlrlHrlrlfHHrlrtrtHrlHHHHrlrlHHrl 




XOOOCOCOCM^G^OlCOCOt-i-I^O^COGOGOfMiOTH^Oi 

OOiOHt'COOOCOOt-iOMHOGOQOt-COCOt-J^COO 
HM^COt-QOCN^Ot-OlHCOlOCCOOlN^OOOOtN 
OOOOOOhhhhhhNiNN!N(NWMWMM^^ 




CDO>OTt*COOK5(M(X)M'tMCOGO»t'OCOiO»0©iOO 

iOi-^CO r JOOCOO(N!NiOOiO!Na)tOt-OMCOQOHTH 

iOHQCiOWCO^MiM-jHt^NOiX)(»HiO^OO(N^CC 

(NiOf-O^t-OTfOOlMCSOiOai^GiOOCKNOO^O^ 

H^CQiOOt-OOHCOTjiCDt-cClC HCOtOOCCOHCOTf 


(0 

fa 
CIS 

>« 










\^ \N NN \^* \N \N \N \N \N \N \N \N 

»-« — (N <N PC *5 T TU5 iCO^OONOOOOaacC- — (N 



26 



INTEREST. 



Table No. 4 gives time in which money will be doubled if 
it is invested either on simple or compound interest, compounded 
annually. 



TABLE No. 4, 



SIMPLE INTEREST. 


COMPOUND INTEREST. 


% 


Years. 


Days. 


% 

2 


Years. 


Days. 


2 


50 




35 


1 


2K 


40 




2K 


28 


30 






33 


120 





23 


162 


%% 


28 


200 


1/ 

'>72 


20 


54 


4 


25 




4 


17 


240 


\% 


22 


80 


4K 


15 


168 


5 


20 




5 


14 


75 





16 


240 


6 


11 


321 


7 


14 


103 


7 


10 


89 


8 


12 


180 


8 


9 


2 


9 


11 


40 


9 


8 


16 


10 


10 




10 


7 


98 


11 


9 


33 


11 


6 


231 


12 


8 


120 


12 


6 


42 



Results of Saving Small Amounts of Honey. 

The following shows how easy it is to accumulate a fortune, 
provided proper steps are taken. 

The table gives the result of daily savings, put in a savings 
bank paying 4 per cent, per year, computed semi-annually : 



Savings 


Savings 


Amount in 


Amount in 


Amount in 


Amount in 


Amount in 


perDay. 


per Mo. 


5 years. 


10 years. 


15 years. 


20 years. 


25 years. 


.05 


$ 1.20 


$ 78.84 


$ 174.96 


$ 292.07 


$ 434.88 


$ 608.94 


.10 


2.40 


157.68 


349.92 


584.15 


869.76 


1,217.88 


.25 


6.00 


394.20 


874.80 


1,460.37 


2,174.40 


3,044.74 


.50 


12.00 


788.40 


1,749.60 


2,920.74 


4,348.80 


6,089.48 


.75 


18.00 


1,182.60 


2,624.40 


4,381.11 


6,523.20 


9,133.22 


$1.00 


24.00 


1,576.80 


3,499.20 


5,841.48 


8,697.60 


12,178.96 



Nearly every person wastes an amount in twenty or thirty 
years, which, if saved and carefully invested, would make a 
family quite independent; but the principle of small savings 
has been lost sight of in the general desire to become wealthy. 



EQUATION OF PAYMENTS 3 . 2*] 

EQUATION OF PAYMENTS. 

When several debts are due at different dates the average 
time when all the debts are due is calculated by the following 
rule: 

Multiply each debt separately by the number of days be- 
tween its own date of maturity and the date of the debt earliest 
due. Divide the sum of these products by the sum of the debts ; 
the quotient will express the number of days subsequent to the 
leading day when the whole debt should be paid in one sum. 

Example. 

A owed to B the following sums: $250 due May 12, $120 
due July 19, $410 due August 16, and $60 due September 21, all 
in the same year. When should the whole sum be paid at once 
in order that neither shall lose any interest? 

Solution : 

May 12 $250 

May 12 to July 19 is 68 days ; 120 X 68 = 8160 
May 12 to Aug. 16 is 96 days; 410 X 96 = 39360 
May 12 to Sept. 21 is 132 days ; 60 X 132 = 7920 



) 55440 = 65.9 

66 days after May 12 will be July 17. 

When several debts are due after different lengths of time, 
the average time is calculated by this rule : Multiply the debt 
by the time ; divide the sum of the products by the sum of the 
debts, and the quotient is the time when all the debts may be 
considered due. 

Example. 
A owed B $600, due in 7 months ; $200 due in one month, and 
$700 due in 3 months. When should the whole debt be paid in 
one sum in order that neither shall lose any interest? 
Solution : 

600 X 7 = 4200 
700 X 3 = 2100 
200 X 1 = 200 

1500 ) 6500 = 4^ months. 

Note : If the debts contain both dollars and cents the 
cents may, if such refinement is required, be considered as deci- 
mal parts of a dollar, but practically in such problems the cents 
may be omitted in the calculation. 



PARTNERSHIP, 



or calculating of proportional parts, is the calculation of the 
parts of a certain quantity in such a way that the ratio between 
the separate parts is equal to the ratio of certain given numbers. 



28 partnership. 

Example 1. 

A composition for welding cast steel consists of 9 parts of 
borax and one part of sal-ammoniac. How much of each, 
borax and sal-ammoniac, must be taken for a mixture of 5 lbs.? 

Solution : 

To X 5 = ±y 2 lbs. borax. 

xV X 5 = Yz lb. sal-ammoniac. 
Example 2. 

An alloy shall consist of 160 parts of copper, 15 parts of tin 
and 5 parts of zinc. How much of each will be used for a cast- 
ing weighing 860 lbs.? 



Solution : 




160 


160 

180 


15 


1 5 

T80 



X 360 = 320 lbs. of copper. 
X 360 = 30 lbs. of tin. 
5 if o X 360 = 10 lbs. of zinc. 

180 

Example 3. 

Four persons — A, B, C and D, are buying a certain amount, 
of goods together. A's part is $500, B's, $100, C's, $250, and 
D's, $150. On the undertaking they are clearing a net profit 
of $120. How much of this is each to have? 

Solution : 

500 A's Part = &% X 120 = $60 

100 B's " = iWo X 120 = 12 

250 C's " = T % 5 o°o X 120 = 30 

150 D's " = T Vo°o X 120 = 18 



$1000 

Example 4. 

Two persons — A and B, are putting money into business, A, 
$2,000 and B, $3,000, but A has his money invested in the busi- 
ness 2 years and B2^ years; the net profit of the undertaking 
is $2,300. How much is each to have of the profit ? 

Solution : 

A, 2000 X 2 = 4000 

B, 3000 X >iy 2 = 7500 

11500 
A's Part is T \%°o% X 2300 = $ 800 
B's " " tVvVo X 2300 = 1500 
In cases like this it must be taken into consideration that 
the time is not equal ; B has not only had the largest capital in- 
vested but he has also had the capital at work in the business 



SQUARE ROOT. 29 

the longest time, namely, 2>£ years, while A has only had his 
capital invested 2 years. The ratio is, therefore, not #2,000 to 
#3,000 but #4,000 to #7,500, because #2,000 in 2 years is equal 
to #4,000 in one year, and #3,000 in 2 l / 2 years is equal to #7,500 in 
one year. 



SQUARE ROOT. 

When the square root is to be extracted the number is di- 
vided into periods consisting of two figures, commencing from 
the extreme right if the number has no decimals, or from the 
decimal point towards the left for the whole numbers and 
towards the right for the decimals. (If the last period of deci- 
mals should have but one figure then annex a cipher, so that 
this period also has two figures, but if the period to the extreme 
left in the integer should happen to have only one figure it 
makes no difference; leave it as it is.) Ascertain the highest 
root of the first period and place it to the right of the number 
as in long division. Square this root and subtract the product 
of this from the first period. To the remainder annex the next 
period of numbers. Take for divisor 20 times the part of the 
root already found* and the quotient is the next figure in the 
root, if the product of this figure and the divisor added to the 
square of the figure does not exceed the dividend. To the 
difference between this sum and the dividend is annexed the 
next period of numbers. For divisor take again 20 times the 
part of the root already found, etc. Continue in this manner 
until the last period is used. If there is any remainder, and a 
more exact root is required, ciphers may be annexed in pairs and 
the operation continued until as many decimals in the root are 
obtained as are wanted. 



Example 1. 

Extract the square root of 271,441 

Solution : a/27 



5 2 = 25 



14 



20 X 5 = 100 ) 214 
100 X 2 + 2 2 = 204 



41 



20 X 52 = 1040 ) 1041 
1040 X 1 + l 2 1041 



521 



0000 
Thus: ^271,441 = 521, because 521 X 521 = 271,441. 

* If this divisor exceeds the dividend, write a cipher in the root; annex the next 
period of numbers, calculate a new divisor, corresponding to the increased root, and 
proceed as explained. 



30 cube root. 

Example 2. 

Extract the square root of i 
Solution : 

\/26j62(56| =5.16 

: : = _' 

20 X 5 = 100 ) 162 
100 X 1 -f 1* = 101 



20 X 51 = 1020 ) 6156 
1020 X 6 -f 6* = 61 56 

~~ 0000 



CUBE ROOT. 

When the cube root is to be extracted, the number is 
divided into periods consisting of three figures. Commencing 
from the extreme right if the number has no decimals, or from 
the decimal point toward the left, for the whole number, and 
toward the right for the decimals. (If the last period of deci- 
mals should not have three figures, then annex ciphers until 
this period also has three figures, but if the period to the 
extreme left in the integer should happen to consist of less than 
three figures it makes no difference : leave it as it is.) Ascer- 
tain highest cube root in the first period and place it to the 
right of the number, the same as in long division. Cube this 
root and subtract the product from the first period. To the 
remainder annex next period of numbers. For the divisor in 
this number take 300 times the square of the part of the root 
already found.* and the quotient is the next figure in the root if 
the product of this figure multiplied by the divisor and added 
to 30 times the part of the root already found, multiplied by the 
square of this quotient and added to the cube of the quotient, 
does not exceed this dividend. To the difference between this 
sum and the dividend is annexed the next period of numbers. 
For divisor take again 300 times the square of the part of the 
root already found, etc. Continue in this manner until the last 
period is used. If there is any remainder from last period, and 
a more exact root is required, ciphers may be annexed three at 
a time, and the operation continued until as many decimals are 
obtained in the root as are wanted. 






CUBE ROOT. 



31 



Example 1. 

Extract the cube root of 275,894,451. 

Solution : 

3 

\^275 
6 3 = 216 



894 



451 



300 X 6 2 = 10800 ) 59894 
10800 X 5 + 30 X X 5 2 + 5 3 = 58625 



300 X 65 2 = 1267500 ) 1269451 
1267500 X 1 + 30 X 65 X l 2 + l 3 == 1269451 



= 651 



0000000 



Thus: 



o 

\/275,894,451 = 651, because 651 X 651 X 651 = 275,894,451, 

Example 2 : 

Extract the cube root of 551.368. 

Solution : 3 

\/~551 

8 3 = 512 

300 X 8 2 = 19200 ) 39368 
19200 X 2 + 30 X 8 X 2 2 + 2 3 = 39368 



368 



S.2 



00000 

The square root of a number consisting of two figures 
will never consist of more than one figure, and the square 
root of a number consisting of four figures will never consist of 
more than two figures ; hence, the rule to divide numbers into 
periods consisting of two figures. 

The cube root of a number consisting of three figures will 
never consist of more than one figure, and the cube root of a 
number consisting of six figures will never consist of more than 
two figures ; hence, the rule to divide the numbers into periods 
consisting of three figures. 

There will always be one decimal in the root for each 
period of decimals in the number of which the root is extracted. 
This relates to both cube and square root. 

The root of a fraction may be found by extracting the 
separate roots of numerator ■ and denominator, or the fraction 
may be first reduced to a decimal fraction before the root is 
extracted. 

The root of a mixed number may be extracted by first re- 
ducing the number to an improper fraction and then extracting 
the separate roots of numerator and denominator, or the number 
may be first reduced to consist of integer and decimal fractions, 
and the root extracted as usual. 



32 CUBE ROOT. 

Radical Quantities Expressed without the Radical Sign. 

The radical sign is not always used in signifying radical 
quantities. Sometimes a quantity expressing a root is written 
as a quantity to be raised into a fractional power. For instance : 

V 16 may be written 165. This is the same value ; thus, 
V^16 = 4 and 162 = 4. 

3 

■V 27 may be written. 27^ = 3. 

3 3 

gl =V8 2 =V til = 4. 

The denominator in the exponent always indicates which 
root is to be extracted. Thus. Ss will be square S and extract 
the cube root from the product. 

Example. 

4 4 

let = V16 3 = V4096 = S. 
Thus, cube 16 and extract the fourth root of the product. 



RECIPROCALS. 



The reciprocal of any number is the quotient which is ob- 
tained when 1 is divided by the number. For instance, the 
reciprocal of 4 is % — 0.25 : the reciprocal of 16 is T x - = 
0.0625, etc. 

Frequently it is a saving of time when performing long 
division to use the reciprocal, as multiplying the dividend by 
the reciprocal of the divisor gives the quotient. For instance, 
divide 4 bv 758. In Table No. 6 the reciprocal of 158 is given 
as 0.0013193. Multiplying 0.0013193 by 4 gives 0.0052772. 
which is correct to sLx decimals. When reducing vulgar frac- 
tions to decimals the reciprocal may be used with advantage. 
For instance, reduce +f to decimals. In Table Xo. 6 the recip- 
rocal of 64 is given as 0.015625, and 15 X 0.015625 = 0.234375. 
which is the decimal of if. 

Important. — Whenever the exact reciprocal is not ex- 
pressible by decimals the result obtained by its use is. as 
explained above, only approximate. 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



TABLE No. 5. Giving Squares, Cubes, Square Roots, 

Cube Roots, and Reciprocals of Fractions and 

flixed Numbers, from ^ to 10. 



n 


^ 2 


n 3 


s/n 


3 

V n 


1 
n 


0.000244 


0.0000038 


0.125 


0.25 


64 


1 

3 2 


0.000977 


0.0000305 


0.17678 


0.31496 


32 


3 

6 4 


0.002196 


0.000103 


0.21651 


0.36056 


21.3333 


1 

16 


0.003900 


0.000244 


0.25 


0.39685 


16 


5 
64 


0.000104 


0.000477 


0.27951 


0.42750 


12.8 


3 
3 2 1 


0.008789 


0.000823 


0.30619 


0.45428 


10.6667 


7 
¥4 


0.011963 


0.001308 


0.33072 


0.47823 


9.1428 


1 

8 


0.015625 


0.001953 


0.35355 


0.5 


8 


9 
^4 ! 


0.01977 


0.00278 


0.375 


0.52002 


7.1111 


5 
3 2 


0.02441 


0.00381 


0.39528 


0.53861 


6.4 


1 1 

3"4 


0.02954 


0.00508 


0.41458 


0.55599 


5.8182 


3 

TS 


0.03516 


0.00659 


0.43301 


0.57236 


5.3333 


1 3 
3"4 


0.04126 


0.00838 


0.45069 


0.58783 


4.9231 


7 


0.04785 


0.01047 


0.46771 


0.60254 


4.5714 


1 5 
3"4 


0.05493 


0.01287 


0.48412 


0.61655 


4.2666 


1 

4 


0.06250 


0.01562 


0.5 


0.62996 


4 


1 7 
3~4 


0.07056 


0.01874 


0.51539 


0.64282 


3.7647 


9 
^2 


0.07910 


0.02225 


0.53033 


0.65519 


3.5556 


1 9 
64 


0.08813 


0.02616 


0.54482 


0.66709 


3.3684 


5 
TF 


0.09766 


0.03052 


0.55902 


0.67860 


3.2 


2 1 

6 4 


0.10766 


0.03533 


0.57282 


0.68973 


3.0476 


1 1 

3 2 


0.11816 


0.04062 


0.58630 


0.70051 


2.9091 


2 3 
F4 


0.12915 


0.04641 


0.59942 


0.71097 


2.7826 


3 

8 


0.14062 


0.05273 


0.61237 


0.72112 


2.6667 


2 5 
64 


0.15258 


0.05960 


0.625 


0.73100 


2.56 


1 3 

32 


0.16504 


0.06705 


0.63738 


0.74062 


2.4615 


2 7 
64 


0.17798 


0.07508 


0.64952 


0.75 


2.3703 


7 
T6" 


0.19141 


0.08374 


0.66144 


0.75915 


2.2857 


2 9 
64 


0.20522 


0.09303 


0.67314 


0.76808 


2.2069 


1 5 

3 2 


0.21973 


0.10300 


0.68465 


0.77681 


2.1333 


3 1 
¥4 


0.23463 


0.11364 


0.69597 


0.78534 


2.0645 


1 

2 


0.25 


0.12500 


0.70711 


0.79370 


2 



34 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


s/~n 


3 

V n 


1 

n 


33 1 
64 


0.26587 


0.13709 


0.71807 


0.80188 


1.9394 


32 


0.28225 


0.14993 


0.728S7 


0.80990 


1.8823 


3 5 
64 


0.29906 


0.16356 


0.73951 


0.81777 


1.8286 


9 
1 6 


0.31641 


0.17789 


0.75000 


0.82548 


1.7778 


3 7 
64 


0.33423 


0.19315 


0.76034 


0.83306 


1.7297 


1 9 
3 2 


0.35254 


0.20932 


0.77055 


0.84049 


1.6842 


3 9 
64 


0.37134 


0.22628 


0.78062 


0.84781 


1.6410 


5 

8 


0.39062 


0.24414 


0.79057 


0.85499 


1.6 


41 
64 


0.41040 


0.26291 


0.80039 


0.86205 


1.5610 


2 1 

3 2 


0.43066 


0.28262 


0.81009 


0.86901 


1.5238 


43 

64 


0.45141 


0.30330 


0.81968 


0.87585 


1.4884 


1 1 

16 


0.47266 


0.32495 


0.82916 


0.88259 


1.4545 


4 5 
64 


0.49438 


0.34761 


0.83853 


0.88922 


1.4222 


23 1 
3 2 


0.51660 


0.37131 


0.84779 


0.89576 


1.3913 


47 
6~4 


0.53931 


0.39605 


0.85696 


0.90221 


1.3617 


3 
4 


0.56250 


0.42187 


0.86603 


0.90856 


1.3333 


49 i 
64 


0.58618 


0.44880 


0.87500 


0.91483 


1.3061 


2 5 i 

"3 2 


0.61035 


0.47684 


0.88388 


0.92101 


1.2800 


5 1 
64 


0.63501 


0.50602 


0.89268 


0.92711 


1.2549 


] 3 
T6 


0.66016 


0.53638 


0.90139 


0.93313 


1.2308 


5 3 

3~4 


0.68579 


0.56792 


0.91001 


0.93907 


1.2075 


2 7 

3 2 


0.71191 


0.60068 


0.91856 


0.94494 


1.1852 


5 5 
64 


0.73853 


0.63467 


0.92702 


0.95074 


1.1636 


7 
¥ 


0.76562 


0.66992 


0.93541 


0.95647 


1.1428 


57 
64 


0.79321 


0.70646 


0.94373 


0.96213 


1.1228 


2 9 

3 2 


0.82129 


0.74429 


0.95197 


0.96772 


1.1034 


59 
'Si 


0.84985 


0.7S346 


0.96014 


0.97325 


1.0847 


1 5 
16 


0.87891 


0.82397 


0.96825 


0.97872 


1.0667 


6 1 
64 


0.90845 


0.86586 


0.97628 


0.98412 


1.0492 


3 1 
32 


0.93848 


0.90915 


0.98425 


0.98947 


1.0323 


63 
64 


0.96899 


0.95385 


0.99216 


0.99476 


1.01587 


1 


1 


1 


1 


1 


1 


ItV 


1.12891 


1.19943 


1.03078 


1.02041 


0.94118 


1.26562 


1.42323 


1.06066 


1.04004 


0.88889 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



35 





n 


n 2 


n 3 


*J n 


V n 


1 

n 


H 

1 5 

1 7 
TIT 


1.41016 


1.67456 


1.08965 


1.05896 


0.84211 


1.5625 


1.953125 


1.11803 


1.07722 


0.8 


1.72266 


2.26099 


1.14564 


1.09488 


0.76190 


1.89062 


2.59961 


1.17260 


1.11199 


0.72727 


2.06641 


2.97046 


1.19896 


1.12859 


0.69565 


H 


2.25 


3.375 


1.22474 


1.14471 


0.66667 


!A 


2.44141 


3.81470 


1.25 


1.16040 


0.64 


Is 

8 


2.640625 


4.29102 


1.27475 


1.17567 


0.61539 


1 1 1 
T6 


2.84766 


4.80542 


1.29904 


1.19055 


0.59260 


1^ 

4 


3.0625 


5.35937 


1.32288 


1.20507 


0.57143 


113 
16 


3.28516 


5.95434 


1.34630 


1.21925 


0.55172 


1 7 

¥ 


3.515625 


6.59180 


1.36931 


1.23311 


0.53333 


115 


3.75391 


7.27319 


1.39194 


1.24666 


0.51613 


2 


4 


8 


1.41421 


1.25992 


0.5 


2 i 

TIT 


4.25390 


8.77368 


1.43614 


1.27291 


0.48485 


21 

8 


4.515625 


9.59582 


1.45774 


1.28564 


0.47059 


2 A 


4.78516 


10.46753 


1.47902 


1.29812 


0.45714 


2 i 


5.0625 


11.390625 


1.5 


1.31037 


0.44444 


2 T 5 6 


5.34766 


12.36646 


1.52069 


1.32239 


0.43243 


2 t 


5.640625 


13.39648 


1.54110 


1.33420 


0.42105 


2 7 
T6" 


5.94141 


14.48217 


1.56125 


1.34580 


0.41026 


21 

2 


6.25 


15.625 


1.58114 


1.35721 


0.4 


2-9- 
16 


6.56541 


16.82641 


1.60078 


1.36843 


0.39024 


2^ 

8 


6.890625 


18.08789 


1.62018 


1.37946 


0.38095 


2 H 


7.22266 


19.41090 


1.63936 


1.39032 


0.37209 


2f 


7.5625 


20.79687 


1.65831 


1.40101 


0.36364 


913 
TF 


7.91016 


22.24731 


1.67705 


1.41155 


0.35555 


27 
¥ 


8.265625 


23.76367 


1.69558 


1.42193 


0.34783 


915 
^T¥ 


8.62891 


25.34724 


1.71391 


1.43216 


0.34042 


3 


9 


27 


1.73205 


1.44225 


0.33333 


H 


9.765625 


30.51758 


1.76777 


1.46201 


0.32 


3i 


10.5625 


34.32812 


1.80278 


1.48125 


0.3077 


3f 


11.390625 


38.44336 


1.83712 


1.5 


0.2963 



36 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n* 


n 3 


V n 


3 

v n 


1 

n 


H 


12.25 


42.875 


1.87083 


1.51829 


0.28571 


H 


13.140625 


47.63476 


1.90394 


1.53616 


0.27586 


3| 


14.0625 


52.73437 


1.93649 


1.55362 


0.26667 


3^ 


15.015625 


58.18555 


1.96850 


1.57069 


0.25806 


4 


16 


64 


2 


1.58740 


0.25 


H 


18.0625 


76.76562 


2.06155 


1.61981 


0.23529 


H 


20.25 


91.125 


2.12132 


1.65096 


0.22222 


*i 


22.5625 


107.17187 


2.17945 


1.68099 


0.21053 


5 


25 


125 


2.23607 


1.70998 


0.2 


«J 


27.5625 


144.70312 


2.291288 


1.73801 


0.19048 


5£ 


30.25 


166.375 


2.34521 


1.76517 


0.18182 


5| 


33.0625 


190.10937 


2.39792 


1.79152 


0.17391 


6 


36 


216 


2.44949 


1.81712 


0.16667 


«i 


39.0625 


244.140625 


2.5 


1.84202 


0.16 


6£ 


42.25 


274.625 


2.54951 


1.86626 


0.15385 


6} 


45.5625 


307.54687 


2.59808 


1.88988 


0.14815 


7 


49 


343 


2.64575 


1.91293 


0.14286 


H 


52.5625 


381.07812 


2.69258 


1.93544 


0.13793 


n 


56.25 


421.875 


2.73861 


1.95743 


0.13333 


"'% 


60.0625 


465.48437 


2.78388 


1.97895 


0.12903 


8 


64 


512 


2.82843 


2 


0.125 


»i 


68.0625 


561.5156 


2.87228 


2.02062 


0.12121 


8J- 


72.25 


614.125 


2.91548 


2.04083 


0.11765 


8| 


76.5625 


669.92187 


2.95804 


2.06064 


0.11428 


9 


81 


729 


3 


2.08008 


0.111111 


91 


85.5625 


791.4531 


3.04138 


2.09917 


0.10811 


9 1 

V 2 


90.25 


857.375 


3.08221 


2.11791 


0.10526 


Q3 
^4 


95.0625 


926.8594 


3.1225 


2.13633 


0.10256 


10 


100 


1000 


3.16228 


2.15443 


0.1 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 37 



TABLE No. 6. Giving Squares, Cubes, Square Roots, 

Cube Roots, and Reciprocals of Numbers from 

o.i to iooo. 



n 


n 2 


n 3 


s/ n 


3 

V n 


1 

n 


0.1 


0.01 


0.001 


0.31623 


0.46416 


10 


0.2 


0.04 


0.008 


0.44721 


0.58480 


5 


0.3 


0.09 


0.027 


0.54772 


0.66943 


O QQQQ 
O.OOOO 


0.4 


0.16 


0.064 


0.63245 


0.73681 


2.5 


0.5 


0.25 - 


0.125 


0.70711 


0.79370 


2 


0.6 


0.36 


0.216 


0.774597 


0.84343 


1.66667 


0.7 


0.49 


0.343 


0.83666 


0.88790 


1.42857 


0.8 


0.68 


0.512 


0.89443 


0.92832 


1.25 


0.9 


0.81 


0.729 


0.94868 


0.96549 


1.11111 


1 


1 


1 


1 


1 


1 


1.1 


1.21 


1.331 


1.04881 


1.03228 


0.909091 


1.2 


1.44 


1.728 


1.09545 


1.06266 


0.833333 


1.3 


1.69 


2.197 


1.14018 


1.09134 


0.769231 


1.4 


1.96 


2.744 


1.18322 


1.11869 


0.714286 


1.5 


2.25 


3.375 


1.22475 


1.14471 


0.666667 


1.6 


2.56 


4.096 


1.26491 


1.16961 


0.625 


1.7 


2.89 


4.913 


1.30384 


1.19347 


0.588235 


1.8 


3.24 


5.832 


1.34164 


1.21644 


0.555556 


1.9 


3.61 


6.859 


1.37840 


1.23855 


0.526316 


2 


4 


8 


1.41421 


1.25992 


0.5 


2.1 


4.41 


9.261 


1.449138 


1.28058 


0.476190 


2.2 


4.84 


10.648 


1.48324 


1 30059 


0.454545 


2.3 


5.29 


12.167 


1.51657 


1.32001 


0.434783 


2.4 


5.76 


13.824 


1.54919 


1.33887 


0.416667 


2.5 


6.25 


15.625 


1.58114 


1.35721 


0.4 


2.6 


6.76 


17.576 


1.61245 


1.37508 


0.384615 


2.7 


7.29 


19.683 


1.64317 


1.39248 


0.37037 


2.8 


7.84 


21.952 


1.67332 


1.40946 


0.357143 


2.9 


8.41 


24.389 


1.70294 


1.42604 


0.344828 


3 


9 


27 


1.73205 


1.44225 


0.333333 


3.2 


10.24 


32.768 


1.78885 


1.47361 


0.3125 


3.4 


11.56 


39.304 


1.84391 


1.50369 


0.294118 


3.6 


12.96 


46.656 


1.89737 


1.53262 


0.277778 


3.8 


14.44 


54.872 


1.94936 


1.56089 


0.263158 


4 


16 


64 


2 


1.58740 


0.25 


4.2 


17.64 


74.088 


2.04939 


1.61343 


0.238095 


4.4 


19.36 


85.184 


2.09762 


1.63868 


0.227273 


4.6 


21.16 


97.336 


2.14476 


1.66310 


0.217391 


4.8 


23.04 


110.592 


2.19089 


1.68687 


0.208333 


5 


25 


I 125 


2.23607 


1.70998 


0.2 



38 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


sf n 


3 

V n 


1 

n 


6 


36 


216 


2.44949 


1.81712 


0.1666667 


7 


49 


343 


2.64575 


1.91293 


0.142857 


8 


64 


512 


2.82843 


2 


0.125 


9 


81 


729 


3 


2.08008 


0.111111 


10 


100 


1000 


3.16228 


2.15443 


0.1 


11 


121 


1331 


3.31662 


2.22398 


0.0909091 


12 


144 


1728 


3.46410 


2.28943 


0.0833333 


13 


169 


2197 


3.60555 


2.35133 


0.0769231 


14 


196 


2744 


3.74166 


2.41014 


0.0714286 


15 


225 


3375 


3.87298 


2.46621 


0.0666667 


16 


256 


4096 


4 


2.51984 


0.0625 


17 


289 


4913 


4.12311 


2.57128 


0.0588235 


18 


324 


5832 


4.24264 


2.62074 


0.0555556 


19 


361 


6859 


4.35890 


2.66840 


0.0526316 


20 


400 


8000 


4.47214 


2.71442 


0.05 


21 


441 


9261 


4.58258 


2.75892 


0.0476190 


22 


484 


10648 


4.69042 


2.80204 


0.0454545 


23 


529 


12167 


4.79583 


2.84387 


0.0434783 


24 


576 


13824 


4.89898 


2.88450 


0.0416667 


25 


625 


15625 


5 


2.92402 


0.04 


26 


676 


17576 


5.09902 


2.96250 


0.0384615 


27 


729 


19683 


5.19615 


3 


0.0370370 


28 


784 


21952 


5.29150 


3.03659 


0.0357143 


29 


841 


24389 


5.38516 


3.07232 


0.0344828 


30 


900 


27000 


5.47723 


3.10723 


0.0333333 


31 


961 


29791 


5.56776 


3.14138 


0.0322581 


32 


1024 


32768 


5.65685 


3.17480 


0.03125 


33 


1089 


35937 


5.74456 


3.20753 


0.0303030 


34 


1156 


39304 


5.83095 


3.23961 


0.0294118 


35 


1225 


42875 


5.91608 


3.27107 


0.0285714 


36 


1296 


46656 


6 


3.30193 


0.0277778 


37 


1369 


50653 


6.08276 


3.33222 


0.0270270 


38 


1444 


54872 


6.16441 


3.36198 


0.0263158 


39 


1521 


59319 


6.245 


3.39121 


0.0256410 


40 


1600 


64000 


6.32456 


3.41995 


0.025 


41 


1681 


68921 


6.40312 


3.44822 


0.0243902 


42 


1764 


74088 


6.48074 


3.47603 


0.0238095 


43 


1849 


79507 


6.55744 


3.50340 


0.0232558 


44 


1936 


85184 


6.63325 


3.53035 


0.0227273 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



39 



n 


n 2 


n 3 


s/n 


3 

s/n 


1 

n 


45 


2025 


91125 


6.70820 


3.55689 


0.0222222 


46 


2116 


97336 


6.78233 


3.58305 


0.0217391 


47 


2209 


103823 


6.85565 


3.60883 


0.0212766 


48 


2304 


110592 


6.92820 


3.63424 


0.0208333 


49 


2401 


117649 


7 


3.65931 


0.0204082 


50 


2500 


125000 


7.07107 


3.68403 


0.02 


51 


2601 


132651 


7.14143 


3.70843 


0.0196078 


52 


2704 


140608 


7.21110 


3.73251 


0.0192308 


53 


2809 


148877 


7.28011 


3.75629 


0.0188679 


54 


2916 


157464 


7.34847 


3.77976 


0.0185185 


55 


3025 


166375 


7.41620 


3.80295 


0.0181818 


56 


3136 


175616 


7.48331 


3.82586 


0.0178571 


57 


3249 


185193 


7.54983 


3.84852 


0.0175439 


58 


3364 


195112 


7.61577 


3.87088 


0.0172414 


59 


3481 


205379 


7.68115 


3.89300 


0.0169492 


60 


3600 


216000 


7.74597 


3.91487 


0.0166667 


61 


3721 


226981 


7.81025 


3.93650 


0.0163934 


62 


3844 


238328 


7.87401 


3.95789 


0.0161290 


63 


3969 


250047 


7.93725 


3.97906 


0.0158730 


64 


4096 


262144 


8 


4 


0.0156250 


65 


4225 


274625 


8.06226 


4.02073 


0.0153846 


66 


4356 


287496 


8.12404 


4.04124 


0.0151515 


67 


4489 


300763 


8.18535 


4.06155 


0.0149254 


68 


4624 


314432 


8.24621 


4.08166 


0.0147059 


69 


4761 


328509 


8.30662 


4.10157 


0.0144928 


70 


4900 


343000 


8.36660 


4.12129 


0.0142857 


71 


5041 


357911 


8.42615 


4. 14082 


0.0140845 


72 


5184 


373248 


8.48528 


4.16017 


0.0138889 


73 


5329 


389017 


8.54400 


4.17934 


0.0136986 


74 


5476 


405224 


8.60233 


4.19834 


0.0135135 


75 


5625 


421875 


8.66025 


4.21716 


0.0133333 


76 


5776 


438976 


8.71780 


4.23582 


0.0131579 


77 


5929 


456533 


8.77496 


4.25432 


0.0129870 


78 


6084 


474552 


8.83176 


4.27266 


0.0128205 


79 


6241 


493039 


8.88819 


4.29084 


0.0126582 


80 


6400 


512000 


8.94427 


4.30887 


0.0125 


81 


6561 


531441 


9 


4.32675 


0.0123457 


82 


6724 


551368 


9.05539 


4.34448 


0.0121951 


83 


6889 


571787 


9.11043 


4.36207 


0.0120482 


84 


7056 


592704 


9.16515 


4.37952 


0.0119048 



40 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


V n 


v n 


1 

n 


85 


7225 


614125 


9.21954 


4.39683 


0.0117647 


86 


7396 


636056 


9.27362 


4.414 


0.0116279 


87 


7569 


658503 


9.32738 


4.43105 


0.0114943 


88 


7744 


681472 


9.38083 


4.44797 


0.0113636 


89 


7921 


704969 


9.43398 


4.46475 


0.0112360 


90 


8100 


729000 


9.48683 


4.48140 


0.0111111 


91 


8281 


753571 


9.53939 


4.49794 


0.0109890 


92 


8464 


778688 


9.59166 


4.51436 


0.0108696 


93 


8649 


804357 


9.64365 


4.53065 


0.0107527 


94 


8836 


830584 


9.69536 


4.54684 


0.0106383 


95 


9025 


857375 


9.74679 


4.56290 


0.0105263 


96 


9216 


884736 


9.79796 


4.57886 


0.0104167 


97 


9409 


912673 


9.84886 


4.59470 


0.0103093 


98 


9604 


941192 


9.89949 


4.61044 


0.0102041 


99 


9801 


970299 


9.94987 


4.62607 


0.0101010 


100 


10000 


1000000 


10 


4.64159 


0.01 


101 


10201 


1030301 


10.04988 


4.65701 


0.0099010 


102 


10404 


1061208 


10.09950 


4.67233 


0.0098039 


103 


10609 


1092727 


10.14889 


4.68755 


0.0097087 


104 


10816 


1124864 


10.19804 


4.70267 


0.0096154 


105 


11025 


1157625 


10.24695 


4.71769 


0.0095238 


106 


11236 


1191016 


10.29563 


4.73262 


0.0094340 


107 


11449 


1225043 


10.34408 


4.74746 


0.0093458 


108 


11664 


1259712 


10.39230 


4.76220 


0.0092593 


109 


11881 


1295029 


10.44031 


4.77686 


0.0091743 


110 


12100 


1331000 


10.4S809 


4.79142 


0.0090909 


111 


12321 


1367631 


10.53565 


4.80590 


0.0090090 


112 


12544 


1404928 


10.58301 


4.82028 


0.0089286 


113 


12769 


1442897 


10.63015 


4.83459 


0.0088496 


114 


12996 


1481544 


10.67708 


4.84881 


0.0087719 


115 


13225 


1520875 


10.72381 


4.86294 


0.0086957 


116 


13456 


1560896 


10.77033 


4.877 


0.0086207 


117 


13689 


1601613 


10.81665 


4.89097 


0.0085470 


118 


13924 


1643032 


10.86278 


4.90487 


0.0084746 


119 


14161 


1685159 


10.90871 


4.91868 


0.0084034 


120 


14400 


1728000 


10.95445 


4.93242 


0.0083333 


121 


14641 


1771561 


11 


4.94609 


0.0082645 


122 


14884 


1815848 


11.04536 


4.95968 


0.0081967 


123 


15129 


1860867 


11.09054 


4.97319 


0.0081301 


124 


15376 


1906624 


11.13553 


4.98663 


0.0080645 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



41 



n 


n 2 


n 3 


v n 


3 

s/ n 


1 

n 


125 


15625 


1953125 


11.18034 


5 


0.008 


126 


15876 


2000376 


11.22497 


5.01330 


0.0079365 


127 


16129 


2048383 


11.26943 


5.02653 


0.0078740 


128 


16384 


2097152 


11.31371 


5.03968 


0.0078125 


129 


16641 


2146689 


11.35782 


5.05277 


0.0077519 


130 


16900 


2197000 


11.40175 


5.06580 


0.0076923 


131 


17161 


2248091 


11.44552 


5.07875 


0.0076336 


132 


17424 


2299968 


11.48913 


5.09164 


0.0075758 


133 


17689 


2352637 


11.53256 


5.10447 


0.0075188 


134 


17956 


2406104 


11.57584 


5.11723 


0.0074627 


135 


18225 


2460375 


11.61895 


5.12993 


0.0074074 


136 


18496 


2515456 


11.66190 


5.14256 


0.0073529 


137 


18769 


2571353 


11.70470 


5.15514 


0.0072993 


138 


19044 


2628072 


11.74734 


5.16765 


0.0072464 


139 


19321 


2685619 


11.78983 


5.18010 


0.0071942 


140 


19600 


2744000 


11.83216 


5.19249 


0.0071429 


141 


19881 


2803221 


11.87434 


5.20483 


0.0070922 


142 


20164 


2863288 


11.91638 


5.21710 


0.0070423 


143 


20449 


2924207 


11.95826 


5.22932 


0.0069930 


144 


20736 


2985984 


12 


5.24148 


0.0069444 


145 


21025 


3048625 


12.04159 


5.25359 


0.0068966 


146 


21316 


3112136 


12.08305 


5.26564 


0.0068493 


147 


21609 


3176523 


12.12436 


5.27763 


0.0068027 


148 


21904 


3241792 


12.16553 


5.28957 


0.0067568 


149 


22201 


3307949 


12.20656 


5.30146 


0.0067114 


150 


22500 


3375000 


12.24745 


5.31329 


0.0066667 


151 


22801 


3442951 


12.28821 


5.32507 


0.0066225 


152 


23104 


3511808 


12.32883 


5.33680 


0.0065789 


153 


23409 


3581577 


12.36932 


5.34848 


0.0065359 


154 


23716 


3652264 


12.40967 


5.36011 


0.0064935 


155 


24025 


3723875 


12.44990 


- 5.37169 


0.0064516 


156 


24336 


3796416 


12.49 


5.38321 


0.0064103 


157 


24649 


3869893 


12.52996 


5.39469 


0.0063694 


158 


24964 


3944312 


12.56981 


5.40612 


0.0063291 


159 


25281 


4019679 


12.60952 


5.41750 


0.0062893 


160 


25600 


4096000 


12.64911 


5.42884 


0.00625 


161 


25921 


4173281 


12.68858 


5.44012 


0.0062112 


162 


26244 


4251528 


12.72792 


5.45136 


0.0061728 


163 


26569 


4330747 


12.76715 


5.46256 


0.0061350 


164 


26896 


4410944 


12.80625 


5.47370 


0.0060976 



42 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


V n 


3 

V n 


1 

n 


165 


27225 


4492125 


12.84523 


5.48481 


0.0060606 


166 


27556 


4574296 


12.88410 


5.49586 


0.0060241 


167 


27889 


4657463 


12.92285 


5.50688 


0.0059880 


168 


28224 


4741632 


12.96148 


5.51785 


0.0059524 


169 


28561 


4826809 


13 


5.52877 


0.0059172 


170 


28900 


4913000 


13.03840 


.5.53966 


0.0058824 


171 


29241 


5000211 


13.07670 


5.55050 


0.0058480 


172 


29584 


5088448 


13.11488 


5.56130 


0.0058140 


173 


29929 


5177717 


13.15295 


5.57205 


0.0057803 


174 


30276 


5268024 


13.19091 


5.58277 


0.0057471 


175 


30625 


5359375 


13.22876 


5.59344 


0.0057143 


176 


30976 


5451776 


13.26650 


5.60408 


0.0056818 


177 


31329 


5545233 


13.30413 


5.61467 


0.0056497 


178 


31684 


5639752 


13.34165 


5.62523 


0.0056180 


179 


32041 


5735339 


13.37909 


5.63574 


0.0055866 


180 


32400 


5832000 


13.41641 


5.64622 


0.0055556 


181 


32761 


5929741 


13.45362 


5.65665 


0.0055249 


182 


33124 


6028568 


13.49074 


5.66705 


0.0054945 


183 


33489 


6128487 


13.52775 


5.67741 


0.0054645 


184 


33856 


6229504 


13.56466 


5.68773 


0.0054348 


185 


34225 


6331625 


13.60147 


5.69802 


0.0054054 


186 


34596 


6434850 


13.63818 


5.70827 


0.0053763 


187 


34969 


6539203 


13.67479 


5.71848 


0.0053476 


1S8 


35334 


6644672 


13.71131 


5.72865 


0.0053191 


189 


35721 


6751269 


13.74773 


5.73879 


0.0052910 


190 


36100 


6859000 


13.78405 


5.74890 


0.0052632 


191 


36481 


6967871 


13.82028 


5.75897 


0.0052356 


192 


36864 


7077888 


13.85641 


5.769 


0.0052083 


193 


37249 


7189057 


13.89244 


5.779 


0.0051813 


194 


37636 


7301384 


13.92839 


5.78896 


0.0051546 


195 


38025 


7414875 


13.96424 


5.79889 


0.0051282 


196 


38416 


7529536 


14 


5.80879 


0.0051020 


197 


38809 


7645372 


14.03567 


5.81865 


0.0050761 


198 


39204 


7762392 


14.07125 


5.82848 


0.0050505 


199 


39601 


7880599 


14.10674 


5.83827 


0.0050251 


200 


40000 


8000000 


14.14214 


5.84804 


0.005 


201 


40401 


8120601 


14.17745 


5.45777 


0.0049751 


202 


40804 


8242408 


14.21267 


5.86747 


0.0049505 


203 


41209 


8365427 


14.24781 


5.87713 


0.0049261 


204 


41616 


8489664 


14.28286 


5.88677 


0.0049020 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



43 



n 


n 2 


n 3 


V n 


3 

V n 


1 

n 


205 


42025 


8615125 


14.31782 


5.89637 


0.0048781 


206 


42436 


8741816 


14.35270 


5.90594 


0.0048544 


207 


42849 


8869743 


14.38749 


5.91548 


0.0048309 


20S 


43264 


8998912 


14.42221 


5.92499 


0.0048077 


209 


43681 


9129320 


14.45683 


5.93447 


0.0047847 


210 


44100 


9261000 


14.49138 


5.94392 


0.0047619 


211 


44521 


9393931 


14.52584 


5.95334 


0.0047393 


212 


44944 


9528128 


14.56022 


5.96273 


0.0047170 


213 


45369 


9663597 


14.59452 


5.97209 


0.0046948 


214 


45796 


9800344 


14.62874 


5.98142 


0.0046729 


215 


46225 


9938375 


14.66288 


5.99073 


0.0046512 


216 


46656 


10077696 


14.69694 


6 


0.0046296 


217 


47089 


10218313 


14.73092 


6.00925 


0.0046083 


218 


47524 


10360232 


14.76482 


6.01846 


0.0045872 


219 


47961 


10503459 


14.79865 


6.02765 


0.0045662 


220 


48400 


10648000 


14.83240 


6.03681 


0.0045455 


221 


48841 


10793861 


14.86607 


6.04594 


0.0045249 


222 


49284 


10941048 


14.89966 


6.05505 


0.0045045 


223 


49729 


11089567 


14.93318 


6.06413 


0.0044843 


224 


50176 


11239424 


14.96663 


6.07318 


0.0044643 


225 


50625 


11390625 


15 


6.08220 


0.0044444 


226 


51076 


11543176 


15.03330 


6.09120 


0.0044248 


227 


51529 


11697083 


15.06652 


6.10017 


0.0044053 


228 


51984 


11852352 


15.09967 


6.10911 


0.0043860 


229 


52441 


12008989 


15.13275 


6.11803 


0.0043668 


230 


52900 


12167000 


15.16575 


6.12693 


0.0043478 


231 


53361 


12326391 


15.19868 


6.13579 


0.0043290 


232 


53824 


12487168 


15.23155 


6.14463 


0.0043103 


233 


54289 


12649337 


15.26434 


6.15345 


0.0042918 


234 


54756 


12812904 


15.29706 


6.16224 


0.0042735 


235 


55225 


12977875 


15.32971 


6.17101 


0.0042553 


236 


55696 


13144256 


15.36229 


6.17975 


0.0042373 


237 


56169 


13312053 


15.39480 


6.18846 


0.0042194 


238 


56644 


13481272 


15.42725 


6.19715 


0.0042017 


239 


57121 


13651919 


15.45962 


6.20582 


0.0041841 


240 


57600 


13824000 


15.49193 


6.21447 


0.0041667 


241 


58081 


13997521 


15.52417 


6.22308 


0.0041494 


242 


58564 


14172488 


15.55635 


6.23168 


0.0041322 


243 


59049 


14348907 


15.58846 


6.24025 


0.0041152 


244 


59536 


14526784 


15.62050 


6.24880 


0.0040984 



44 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n s 


V^T 


3 


1 

n 


245 


60025 


14706125 


15.65248 


6.25732 


0.0040816 


246 


60516 


14886936 


15.68439 


6.26583 


0.0040650 


247 


61009 


15069223 


15.71623 


6.27431 


0.0040486 


248 


61504 


15252992 


15.74802 


6.28276 


0.0040323 


249 


62001 


15438249 


15.77973 


6.29119 


0.0040161 


250 


62500 


15625000 


15.81139 


6.29961 


0.004 


251 


63001 


15813251 


15.84298 


6.30799 


0.0039841 


252 


63504 


16003008 


15.87451 


6.31636 


0.0039683 


253 


64009 


16194277 


15.90597 


6.32470 


0.0039526 


254 


64516 


16387064 


15.93738 


6.33303 


0.0039370 


255 


65025 


16581375 


15.96872 


6.34133 


0.0039216 


256 


65536 


16777216 


16 


6.34960 


0.0039062 


257 


66049 


16974593 


16.03122 


6.35786 


0.0038911 


258 


66564 


17173512 


16.06238 


6.36610 


0.0038760 


259 


67081 


17373979 


16.09348 


6.37431 


0.0038610 


260 


67600 


17576000 


16.12452 


6.38250 


0.0038462 


261 


68121 


17779581 


16.15549 


6.39068 


0.0038314 


262 


68644 


17984728 


16.18641 


6.39883 


0.0038168 


263 


69169 


18191447 


16.21727 


6.40696 


0.0038023 


264 


69696 


18399744 


16.24808 


6.41507 


0.0037879 


265 


70225 


18609625 


16.27882 


6.42316 


0.0037736 


266 


70756 


18821096 


16.30951 


6.43123 


0.0037594 


267 ! 


71289 


19034163 


16.34013 


6.43928 


0.0037453 


268 


71824 


19248832 


16.37071 


6.44731 


0.0037313 


269 


72361 


19465109 


16.40122 


6.45531 


0.0037175 


270 


72900 


19683000 


16.43168 


6.46330 


0.0037037 


271 


73441 


19902511 


16.46208 


6.47127 


0.00369 


272 


73984 


20123648 


16.49242 


6.47922 


0.0036765 


273 


74529 


20346417 


16.52271 


6.48715 


0.0036630 


274 


75076 


20570824 


16.55295 


6.49507 


0.0036496 


275 


75625 


20796875 


16.58312 


6.50296 


0.0036364 


276 


76176 


21024576 


16.61325 


6.51083 


0.0036232 


277 


76729 


21253933 


16.64332 


6.51868 


0.0036101 


278 


77284 


21484952 


16.67333 


6.52652 


0.0035971 


279 


77841 


21717639 


16.70329 


6.53434 


0.0035842 


280 


78400 


21952000 


16.73320 


6.54213 


0.0035714 


281 


78961 


22188041 


16.76305 


6.54991 


0.0035587 


282 


79524 


22425768 


16.79286 


6.55767 


0.0035461 


283 


80089 


22665187 


16.82260 


6.56541 


0.0035336 


284 


80656 


22906304 


16.85230 


6.57314 


0.0035211 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



45 



n 


n 2 


n 3 


V n 


v n 


1 

n 


285 


81225 


23149125 


16.88194 


6.58084 


0.0035088 


286 


81796 


23393656 


16.91153 


6.58853 


0.0034965 


287 


82369 


23639903 


16.94107 


6.59620 


0.0034843 


288 


82944 


23887872 


16.97056 


6.60385 


0.0034722 


289 


83521 


24137569 


17 


6.61149 


0.0034602 


290 


84100 


24389000 


17.02939 


6.61911 


0.0034483 


291 


84681 


24642171 


17.05872 


6.62671 


0.0034364 


292 


85264 


24897088 


17.08801 


6.63429 


0.0034247 


293 


85849 


25153757 


17.11724 


6.64185 


0.0034130 


294 


86436 


25412184 


17.14643 


6.64940 


0.0034014 


295 


87025 


25672375 


17.17556 


6.65693 


0.0033898 


296 


87616 


25934336 


17.20465 


6.66444 


0.0033784 


297 


88209 


26198073 


17.23369 


6.67L94 


0.0033670 


298 


88804 


26463592 


17.26268 


6.67942 


0.0033557 


299 


89401 


26730899 


17.29162 


6.68688 


0.0033445 


300 


90000 


27000000 


17.32051 


6.69433 


0.0033333 


301 


90601 


27270901 


17.34935 


6.70176 


0.0033223 


302 


91204 


27543608 


17.37815 


6.70917 


0.0033113 


303 


91809 


27818127 


17.40690 


6.71657 


0.0033003 


304 


92416 


28094464 


17.43560 


6.72395 


0.0032895 


305 


93025 


28372625 


17.46425 


6.73132 


0.0032787 


306 


93636 


28652616 


17.49286 


6.73866 


0.0032680 


307 


94249 


28934443 


17.52142 


6.746 


0.0032573 


308 


94864 


29218112 


17.54993 


6.75331 


0.0032468 


309 


95481 


29503629 


17.57840 


6.76061 


0.0032362 


310 


96100 


29791000 


17.60682 


6.76790 


0.0032258 


311 


96721 


30080231 


17.63519 


6.77517 


0.0032154 


312 


97344 


30371328 


17.66352 


6.78242 


0.0032051 


313 


97969 


30664297 


17.69181 


6.78966 


0.0031949 


314 


98596 


30959144 


17.72005 


6.79688 


0.0031847 


315 


99225 


31255875 


17.74824 


6.80409 


0.0031746 


316 


99856 


31554496 


17.77639 


6.81128 


0.0031646 


317 


100489 


31855013 


17.80449 


6.81846 


0.0031546 


318 


101124 


32157432 


17.83255 


6.82562 


0.0031447 


319 


101761 


32461759 


17.86057 


6.83277 


0.0031348 


320 


102400 


32768000 


17.88854 


6.83990 


0.0031250 


321 


103041 


33076161 


17.91647 


6.84702 


0.0031153 


322 


103684 


33386248 


17.94436 


6.85412 


0.0031056 


323 


104329 


33698267 


17.97220 


6.86121 


0.0030960 


324 


104976 


34012224 


18 


6.86829 


0.0030864 



46 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



71 


ft 2 


n 3 


Vn 


3 

srn 


1 

n 


325 


105625 


34328125 


18.02776 


6.87534 


0.0030769 


326 


106276 


34645976 


18.05547 


6.88239 


0.0030675 


327 


106929 


34965783 


18.08314 


6.88942 


0.0030581 


32S 


107584 


35287552 


18.11077 


6.89643 


0.0030488 


329 


108241 i 


35611289 


18.13836 


6.90344 


0.0030395 


330 


108900 l 


35937000 


18.16590 


6.91042 


0.0030303 


331 


109561 


36264691 


18.19341 


6.91740 


0.0030211 


332 


110224 


36594368 


1S.22087 


6.92436 


0.0030120 


333 


110889 ; 


36926037 


18.24S29 


6.93131 


0.0030030 


334 


111556 


37259704 


18.27567 


6.93823 


0.0029940 


335 


112225 


37595375 


18.30301 


6.94515 


0.0029851 


336 


112896 


37933056 


18.33030 


6.95205 


0.0029762 


337 


113569 


38272753 


18.35756 


6.95894 


0.0029674 


338 


114244 


386144*72 


IS. 38478 


6.96582 


0.0029586 


339 


114921 


38958219 


18.41195 


6.97268 


0.0029499 


340 


115600 


39304000 


18.43909 


6.97953 


0.0029412 


341 


116281 


39651821 


18.46619 


6.98637 


0.0029326 


342 


116964 


40001688 


18.49324 


6.99319 


0.0029240 


343 


117649 


40353607 


18.52026 


7 


0.0029155 


344 


118336 


40707584 


18.54724 


7.00680 


0.0029070 


345 


119025 


41063625 


18.57418 


7.01358 


0.0028986 


346 


• 119716 


41421736 


18.60108 


7.02035 


0.0028902 


347 


120409 


41781923 


18.62794 


7.02711 


0.0028818 


348 


121104 


42144192 


18.65476 


7.03385 


0.0028736 


349 


121801 


42508549 


18.68154 


7.04059 


0.0028653 


350 


122500 


42875000 


18.70829 


7.04730 


0.0028571 


351 


123201 


43243551 


18.73499 


7.054 


0.0028490 


352 


123904 


43614208 


18.76166 


7.06070 


0.0028409 


353 


124609 


43986977 


18.78829 


7.06738 


0.002S329 


354 


125316 


44361864 


18.81489 


7.07404 


0.0028249 


355 


126025 


44738875 


18.84144 


7.08070 


0.0028169 


356 


126736 


45118016 


18.86796 


7.08734 


0.0028090 


357 


127449 


45499293 


18.89444 


7.09397 


0.0028011 


358 


128164 


45882712 


18.92089 


7.10059 


0.0027933 


359 


128881 


46268279 


18.94730 


7.10719 


0.0027855 


360 


129600 


46656000 


IS. 97367 


7.11379 


0.0027778 


361 


130321 


47045881 


19 


7.12037 


0.0027701 


362 


131044 


47437928 


19.02630 


7.12694 


0.0027624 


363 


131769 


47832147 


19.05256 


7.13349 


0.0027548 


364 


132496 


48228544 


19.07878 


7.14004 


0.0027473 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



47 



n 


n 2 


n 3 


\/~n 


3 

v n 


1 

n 


365 


133225 


48627125 


19.10497 


7.14657 


0.0027397 


366 


133956 


49027896 


19.13113 


7.15309 


0.0027322 


367 


134689 


49430863 


19.15724 


7.15960 


0.0027248 


368 


135424 


49836032 


19.18333 


7.16610 


0.0027174 


369 


136161 


50243409 


19.20937 


7.17258 


0.0027100 


370 


136900 


50653000 


19.23538 


7.17905 


0.0027027 


371 


137641 


51064811 


19.26136 


7.18552 


0.0026954 


372 


138384 


51478848 


19.28730 


7.19197 


0.0026882 


373 


139129 


51895117 


19.31321 


7.19841 


0.0026810 


374 


139876 


52313624 


19.33908 


7.20483 


0.0026738 


375 


140625 


52734375 


19.36492 


7.21125 


0.0026667 


376 


141376 


53157376 


19.39072 


7.21765 


0.0026596 


377 


142129 


53582633 


19.41649 


7.22405 


0.0026525 


378 


142884 


54010152 


19.44222 


7.23643 


0.0026455 


379 


143641 


54439939 


19.46792 


7.23680 


0.0026385 


380 


144400 


54872000 


19.49359 


7.24316 


0.0026316 


381 


145161 


55306341 


19.51922 


7.24950 


0.0026247 


382 


145924 


55742968 


19.54482 


7.25584 


0.0026178 


383 


146689 


56181887 


19.57039 


7.26217 


0.0026110 


384 


147456 


56623104 


19.59592 


7.26848 


0.0026042 


385 


148225 


57066625 


19.62142 


7.27479 


0.0025974 


3S6 


148996 


57512456 


19.64688 


7. 281 OS 


0.0025907 


387 


149769 


57960603 


19.67232 


7.28736 


0.0025840 


388 


150544 


58411072 


19.69772 


7.29363 


0.0025773 


389 


151321 


58863869 


19.72308 


7.29989 


0.0025707 


390 


152100 


59319000 


19.74842 


7.30614 


0.0025641 


391 


152881 


59776471 


19.77372 


7.31238 


0.0025575 


392 


153664 


60236288 


19.79899 


7.31861 


0.0025510 


393 


154449 


60698457 


19.82423 


7.32483 


0.0025445 


394 


155236 


61162984 


19.84943 


7.33104 


0.0025381 


395 


156025 


61629875 


19.87461 


7.33723 


0.0025316 


396 


156816 


62099136 


19.89975 


7.34342 


0.0025253 


397 


157609 


62570773 


19.92486 


7.34960 


0.0025189 


398 


15S404 


63044792 


19.94994 


7.35576 


0.0025126 


399 


159201 


63521199 


19.97498 


7.36192 


0.0025063 


400 


160000 


64000000 


20 


7.36806 


0.0025 


401 


160801 


64481201 


20 02498 


7.37420 


0.0024938 


402 


161604 


64964808 


20.04994 


7.38032 


0.0024876 


403 


162409 


65450827 


20.07486 


7.38644 


0.0024814 


404 


163216 


65939264 


20.09975 


7.39254 


0.0024752 



4 8 



SQUARES, CUBES. ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


V n 


3 


1 

n 


405 


164025 


66430125 


20.12461 


7.39864 


0.0024691 


406 


164836 


66923416 


20.14944 


7.40472 


0.0024631 


407 


165649 


67419143 


20.17424 


7.41080 


0.0024570 


408 


166464 


67917312 


20.19901 


7.41686 


0.0024510 


409 


167281 


68417929 


20.22375 


7.42291 


0.0024450 


410 


168100 


68921000 


20.24846 


7.42896 


0.0024390 


411 


168921 


69426531 


20.27313 


7.43499 


0.0024331 


412 


169744 


69934528 


20.29778 


7.44102 


0.0024272 


413 


170569 


70444997 


20.32240 


7.44703 


0.0024213 


414 


171396 


70957944 


20.34699 


7.45304 


0.0024155 


415 


172225 


71473375 


20.37155 


7.45904 


0.0024096 


416 


173056 


71991296 


20.39608 


7.46502 


0.0024038 


417 


173889 


72511713 


20.42058 


7.471 


0.0023981 


418 


174724 


73034632 


20.44505 


7.47697 


0.0023923 


419 


175561 


73560059 


20.46949 


7.48292 


0.0023866 


420 


176400 


74088000 


20.49390 


7.48887 


0.0023810 


421 


177241 


74618461 


20.51828 


7.49481 


0.0023753 


422 


178084 


75151448 


20.54264 


7.50074 


0.0023697 


423 


178929 


75686967 


20.56696 


7.50666 


0.0023641 


424 


179776 


76225024 


20.59126 


7.51257 


0.0023585 


425 


180625 


76765625 


20.61553 


7.51847 


0.0023529 


426 


181476 


7730S776 


20.63977 


7.52437 


0.0023474 


427 


182329 


77854483 


20.66398 


7.53025 


0.0023419 


428 


183184 


78402752 


20.68816 


7.53612 


0.0023364 


429 


184041 


78953589 


20.71232 


7.54199 


0.0023310 


430 


184900 


79507000 


20.73644 


7.54784 


0.0023256 


431 


185761 


80062991 


20.76054 


7.55369 


0.0023202 


432 


186624 


80621568 


20.78461 


7.55953 


0.0023148 


433 


187489 


81182737 


20.80865 


7.56535 


0.0023095 


434 


188356 


81746504 


20.83267 


7.57117 


0.0023041 


435 


189225 


82312875 


20.85665 


7.57698 


0.0022989 


436 


190096 


82881856 


20.88061 


7.58279 


0.0022936 


437 


190969 


83453453 


20.90455 


7.58858 


0.0022883 


438 


191844 


84027672 


20.92845 


7.59436 


0.0022831 


439 


192721 


84604519 


20.95233 


7.60014 


0.0022779 


440 


193600 


85184000 


20.97618 


7.60590 


0.0022727 


441 


194481 


85766121 


21 


7.61166 


0.0022676 


442 


195364 


86350888 


21.02380 


7.61741 


0.0022624 


443 


196249 


86938307 


21.04757 


7.62315 


0.0022573 


444 


197136 


87528384 


21.07131 


7.62888 


0.0022523 

. . 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



49 



n 


n 2 


n 3 


V n 


3 

V n 


1 

n 


445 


198025 


88121125 


21.09502 


7.63461 


0.0022472 


44(3 


198916 


88716536 


21.11871 


7.64032 


0.0022422 


447 


199809 


89314623 


21.14237 


7.64603 


0.0022371 


448 


200704 


89915392 


21.16601 


7.65172 


0.0022321 


449 


201601 


90518849 


21.18962 


7.65741 


0.0022272 


450 


202500 


91125000 


21.21320 


7.66309 


0.0022222 


451 


203401 


91733851 


21.23676 


7.66877 


0.0022173 


452 


204304 


92345408 


21.26029 


7.67443 


0.0022124 


453 


205209 


92959677 


21.28380 


7.68009 


0.0022075 


454 


206116 


93576664 


21.30728 


7.68573 


0.0022026 


455 


207025 


94196375 


21.33073 


7.69137 


0.0021978 


456 


207936 


94818816 


21.35416 


7.69700 


0.0021930 


457 


208849 


95443993 


21.37756 


7.70262 


0:002 1882 


458 


209764 


96071912 


21.40093 


7.70824 ' 


0.0021834 


459 


210681 


96702579 


21.42429 


7.71384 


0.0021786 


460 


211600 


97336000 


21.44761 


7.71944 


0.0021739 


461 


212521 


97972181 


21.47091 


7.72503 


0.0021692 


462 


213444 


98611128 


21.49419 


7.73061 


0.0021645 


463 


214369 


99252847 


21.51743 


7.73619 


0.0021598 


464 


215296 


99897344 


21.54066 


7.74175 


0.0021552 


465 


216225 


100544625 


21.56386 


7.74731 


0.0021505 


466 


217156 


101194696 


21.58703 


7.75286 


0.0021459 


467 


218089 


101847563 


21.61018 


7.75840 


0.0021413 


468 


219024 


102503232 


21.63331 


7.76394 


0.0021368 


469 


219961 


103161709 


21.65641 


7.76946 


0.0021322 


470 


220900 


103823000 


21.67948 


7.77498 


0.0021277 


471 


221841 


104487111 


21.70253 


7.78049 


0.0021231 


472 


222784 


105154048 


21.72556 


7.78599 


0.0021186 


473 


223729 


105823817 


21.74856 


7.79149 


0.0021142 


474 


224676 


106496424 


21.77154 


7.79697 


0.0021097 


475 


225625 


107171875 


21.79449 


7.80245 


0.0021053 


476 


226576 


107850176 


21.81742 


7.80793 


0.0021008 


477 


227529 


108531333 


21.84033 


7.81339 


0.0020965 


478 


228484 


109215352 


21.86321 


7.81885 


0.0020921 


479 


229441 


109902239 


21.88607 


7.82429 


0.0020877 


480 


230400 


110592000 


21.90890 


7.82974 


0.0020833 


481 


231361 


111284641 


21.93171 


7.83517 


0.0020790 


482 


232324 


111980168 


21.95450 


7.84059 


0.0020747 


483 


233289 


112678587 


21.97726 


7.84601 


0.0020704 


484 


234256 


113379904 


22 


7.85142 


0.0020661 



So 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


vST 


V n 


1 

n 


485 


235225 


114084125 


22.02272 


7.85683 


0.0020619 


486 


236196 


114791256 


22.04541 


7.86222 


0.0020576 


487 


237169 


115501303 


22.06808 


7.86761 


0.0020534 


488 


238144 


116214272 


22.09072 


7.87299 


0.0020492 


489 


239121 


116930169 


22.11334 


7.87837 


0.0020450 


490 


240100 


117649000 


22.13594 


7.88374 


0.0020408 


491 


241081 


118370771 


22.15852 


7.88909 


0.0020367 


492 


242064 


119095488 


22.18107 


7.89445 


0.0020325 


493 


243049 


119823157 


22,20360 


7.89979 


0.0020284 


494 


244036 


120553784 


22.22611 


7.90513 


0.0020243 


495 


245025 


121287375 


22.24860 


7.91046 


0.0020202 


496 


246016 


122023936 


22.27106 


7.91578 


0.0020161 


497 


247009 


122763473 


22.29350 


7.92110 


0.0020121 


498 


248004 


123505992 


22.31591 


7.92641 


0.0020080 


499 


249001 


124251499 


22.33831 


7.93171 


0.0020040 


500 


250000 


125000000 


22.36068 


7.93701 


0.002 


501 


251001 


125751501 


22.38303 


7.94229 


0.0019960 


502 


252004 


126506008 


22.40536 


7.94757 


0.0019920 


503 


253009 


127263527 


22.42766 


7.95285 


0.0019881 


504 


254016 


128024064 


22.44994 


7.95811 


0.0019841 


505 


255025 


128787626 


22.47221 


7.96337 


0.0019802 


506 


256036 


129554216 


22.49444 


7.96863 


0.0019763 


507 


257049 


130323843 


22.51666 


7.97387 


0.0019724 


508 


258064 


131096512 


22.53886 


7.97911 


0.0019685 


509 


259081 


131872229 


22.56103 


7.98434 


0.0019646 


510 


260100 


132651000 


22.58318 


7.98957 


0.0019608 


511 


261121 


133432831 


22.60531 


7.99479 


0.0019569 


512 


262144 


134217728 


22.62742 


8 


0.0019531 


513 


263169 


135005697 


22.64950 


8.00520 


0.0019493 


514 


264196 


135796744 


22.67157 


8.01040 


0.0019455 


515 


265225 


136590875 


22.69361 


8.01559 


0.0019417 


516 


266256 


137388096 


22.71563 


8.02078 


0.0019380 


517 


267289 


138188413 


22.73763 


8.02596 


0.0019342 


518 


268324 


138991832 


22.75961 


8.03113 


0.0019205 


519 


269361 


139798359 


22.78157 


8.03629 


0.0019268 


520 


270400 


140608000 


22.80351 


8.04145 


0.0019231 


521 


271441 


141420761 


22.82542 


8.04660 


0.0019194 


522 


272484 


142236648 


22.84732 


8.05175 


0.0019157 


523 


273529 


143055667 


22.86919 


8.05689 


0.0019120 


524 


274576 


143877824 


22.89105 


8.06202 


0.0019084 



Squares, cubes, roots, and reciprocals. 



51 



n 


ft 2 


n d 


V n 


3 

V n 


1 

n 


525 


275625 


144703125 


22.91288 


8.06714 


0.0019048 


526 


276676 


145531576 


22.93469 


8.07226 


0.0019011 


527 


277729 


146363183 


22.95648 


8.07737 


0.0018975 


528 


278784 


147197952 


22.97825 


8.08248 


0.0018939 


529 


279841 


148035889 


23 


8.08758 


0.0018904 


530 


280900 


148877000 


23.02173 


8.09267 


0.0018868 


531 


281961 


149721291 


23.04344 


8.09776 


0.0018832 


532 


283024 


150568768 


23.06513 


8.10284 


0.0018797 


533 


284089 


151419437 


23.08679 


8.10791 


0.0018762 


534 


285156 


152273304 


23.10844 


8.11298 


0.0018727 


535 


286225 


153130375 


23.13007 


8.11804 


0.0018692 


536 


287296 


153990656 


23.15167 


8.12310 


0.0018657 


537 


288369 


154854153 


23.17326 


8.12814 


0.0018622 


538 


289444 


155720872 


23.19483 


8.13319 


0.0018587 


539 


290521 


156590819 


23.21637 


8.13822 


0.0018553 


540 


291600 


157464000 


23.23790 


8.14325 


0.0018519 


541 


292681 


158340421 


23.25941 


8.14828 


0.0018484 


542 


293764 


159220088 


23.28089 


8.15329 


0.0018450 


543 


294849 


160103007 


23.30236 


8.15831 


0.0018416 


544 


295936 


160989184 


23.32381 


8.16331 


0.0018382 


545 


297025 


161878625 


23.34524 


8.16831 


0.0018349 


546 


298116 


162771336 


23.36664 


8.17330 


0.0018315 


547 


299209 


163667323 


23.38803 


8.17829 


0.0018282 


548 


300304 


164566592 


23.40940 


8.18327 


0.0018248 


549 


301401 


165469149 


23.43075 


8.18824 


0.0018215 


550 


302500 


166375000 


23.45208 


8.19321 


0.0018182 


551 


303601 


167284151 


23.47339 


8.19818 


0.0018149 


552 


304704 


168196608 


23.49468 


8.20313 


0.0018116 


553 


305809 


169112377 


23.51595 


8.20808 


0.0018083 


554 


306916 


170031464 


23.53720 


8.21303 


0.0018051 


555 


308025 


170953875 


23.55844 


8.21797 


0.0018018 


556 


309136 


171879616 


23.57965 


8.22290 


0.0017986 


557 


310249 


172808693 


23.60085 


8.22783 


0.0017953 


558 


311364 


173741112 


23.62202 


8.23275 


0.0017921 


559 


312481 


174676879 


23.64318 


8.23766 


0.0017889 


560 


313600 


175616000 


23.66432 


8.24257 


0.0017857 


561 


314721 


176558481 


23.68544 


8.24747 


0.0017825 


562 


315844 


177504328 


23.70654 


8.25237 


0.0017794 


563 


316969 


178453547 


23.72762 


8.25726 


0.0017762 


564 


318096 


179406144 


23.74868 


8.26215 


0.0017730 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n* 


n 3 


V n 


3 

V n 


1 

n 


565 


319225 


180362125 


23.76973 


8.26703 


0.0017699 


566 


320356 


181321496 


23.79075 


8.27190 


0.0017668 


567 


321489 


182284263 


23.81176 


8.27677 


0.0017637 


568 


322624 


183250432 


23.83275 


8.28163 


0.0017606 


569 


323761 


184220009 


23.85372 


8.28649 


0.0017575 


570 


324900 


185193000 


23.87467 


8.29134 


0.0017544 


571 


326041 


186169411 


23.89561 


8.29619 


0.0017513 


572 


327184 


187149248 


23.91652 


8.30103 


0.0017483 


573 


328329 


188132517 


23.93742 


8.30587 


0.0017452 


574 


329476 


189119224 


23.95830 


8.31069 


0.0017422 


575 


330625 


190109375 


23.97916 


8.31552 


0.0017391 


576 


331776 


191102976 


24 


8.32034 


0.0017361 


577 


332929 


192100033 


24.02082 


8.32515 


0.0017331 


578 


334084 


193100552 


24.04163 


8.32995 


0.0017301 


579 


335241 


194104539 


24.06242 


8.33476 


0.0017271 


580 


336400 


195112000 


24.08319 


8.33955 


0.0017241 


581 


337561 


196122941 


24.10394 


8.34434, 


0.0017212 


582 


338724 


197137368 


24.12468 


8.34913 


0.0017182 


583 


339889 


198155287 


24.14539 


8.35390 


0.0017153 


584 


341056 


199176704 


24.16609 


8.35868 


0.0017123 


585 


342225 


200201625 


24.18677 


8.36345 


0.0017094 


586 


343396 


201230056 


24.20744 


8.36821 


0.0017065 


587 


344569 


202262003 


24.22808 


8.37297 


0.0017036 


588 


345744 


203297472 


24.24871 


8.37772 


0.0017007 


589 


346921 


204336469 


24.26932 


8.38247 


0.0016978 


590 


348100 


205379000 


24.28992 


8.38721 


0.0016949 


591 


349281 


206425071 


24.31049 


8.39194 


0.0016920 


592 


350464 


207474688 


24.33105 


8.39667 


0.0016892 


593 


351649 


208527857 


24.35159 


8.40140 


0.0016863 


594 


352836 


209584584 


24.37212 


8.40612 


0.0016835 


595 


354025 


210644875 


24.39262 


8.41083 


0.0016807 


596 


355216 


211708736 


24.41311 


8.41554 


0.0016779 


597 


356409 


212776173 


24.43358 


8.42025 


0.0016750 


598 


357604 


213847192 


24.45404 


8.42494 


0.0016722 


599 


358801 


214921799 


24.47448 


8.42964 


0.0016694 


600 


360000 


216000000 


24.49490 


8.43433 


0.0016667 


601 


361201 


217081801 


24.51530 


8.43901 


0.0016639 


602 


362404 


218167208 


24.53569 


8.44369 


0.0016611 


603 


363609 


219256227 


24.55606 


8.44836 


0.0016584 


604 


364816 


220348864 


24.57641 


8.45303 


0.0016556 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



53 



n 


n 2 


n 3 


V//~ 


3 . 

V r n 


1 

n 


605 


366025 


221445125 


24.59675 


8.45769 


0.0016529 


606 


367236 


222545016 


24.61707 


8.46235 


0.0016502 


607 


368449 


223648543 


24.63737 


8.46700 


0.0016474 


608 


369664 


224755712 


24.65766 


8.47165 


0.0016447 


609 


370881 


225866529 


24.67793 


8.47629 


0.0016420 


610 


372100 


226981000 


24.69818 


8.48093 


0.0016393 


611 


373321 


228099131 


24.71841 


8.48556 


0.0016367 


612 


374544 


229220928 


24.73863 


8.49018 


0.0016340 


613 


375769 


230346397 


24.75884 


8.49481 


0.0016313 


614 


376996 


231475544 


24.77902 


8.49942 


0.0016287 


615 


378225 


232608375 


24.79919 


8.50404 


0.0016260 


616 


379456 


233744896 


24.81935 


S. 50864 


0.0016234 


617 


380689 


234885113 


24.83948 


8.51324 


0.0016207 


618 


381924 


236029032 


24.85961 


8.51784 


0.0016181 


619 


383161 


237176659 


24.87971 


8.52243 


0.0016155 


620 


384400 


238328000 


24.89980 


8.52702 


0.0016129 


621 


385641 


239483061 


24.91987 


8.53160 


0.0016103 


622 


386884 


240641848 


24.93993 


8.53618 


0.0016077 


623 


388129 


241804367 


24.95997 


8.54075 


0.0016051 


624 


389376 


242970624 


24.97999 


8.54532 


0.0016026 


625 


390625 


244140625 


25 


8.54988 


0.0016000 


626 


391876 


245314376 


25.01999 


8.55444 


0.0015974 


627 


393129 


246491883 


25.03997 


8.55899 


0.0015949 


628 


394384 


247673152 


25.05993 


8.56354 


0.0015924 


629 


395641 


248858189 


25.07987 


8.56808 


0.0015898 


630 


396900 


250047000 


25.09980 


8.57262 


0.0015873 


631 


398161 


251239591 


25.11971 


8.57715 


0.0015848 


632 


399424 


252435968 


25.13961 


8.58168 


0.0015823 


633 


400689 


253636137 


25.15949 


8.58622 


0.0015798 


634 


401956 


254840104 


25.17936 


8.59072 


0.0015773 


635 


403225 


256047875 


25.19921 


8.59524 


0.0015748 


636 


404496 


257259456 


25.21904 


8.59975 


0.0015723 


637 


405769 


258474853 


25.23886 


8.60425 


0.0015699 


638 


407044 


259694072 


25.25866 


8.60875 


0.0015674 


639 


408321 


260917119 


25.27845 


8.61325 


0.0015649 


640 


409600 


262144000 


25.29822 


8.61774 


0.0015625 


641 


410881 


263374721 


25.31798 


8.62222 


0.0015601 


642 


412164 


264609288 


25.33772 


8.62671 


0.0015576 


643 


413449 


265847707 


25.35744 


8.63118 


0.0015552 


644 


414736 


267089984 


25.37716 


8.63566 


0.0015528 



54 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


sf n 


3 

s/ n 


1 

n 


645 


416025 


268336125 


25.39685 


8.64012 


0.0015504 


646 


417316 


269586136 


25.41653 


8.64459 


0.0015480 


647 


418609 


270840023 


25.43619 


8.64904 


0.0015456 


648 


419904 


272097792 


25.45584 


8.65350 


0.0015432 


649 


421201 


273359449 


35.47548 


8.65795 


0.0015408 


650 


422500 


274625000 


25,49510 


8.66239 


0.0015385 


651 


423801 


275894451 


25.51470 


8.66683 


0.0015361 


652 


425104 


2771 67808 


25.53429 


8.67127 


0.0015337 


653 


426409 


278445077 


25.55386 


8.67570 


0.0015314 


654 


427716 


279726264 


25.57342 


8.68012 


0.0015291 


655 


429025 


2X1011375 


25.59297 


8.68455 


0.0015267 


656 


430336 


282300416 


25.61250 


8.68896 


0.0015244 


657 


431649 


283593393 


25.63201 


8.69338 


0.0015221 


658 


432964 


284890312 


25.65151 


8.69778 


0.0015198 


659 


434281 


286191179 


25.67100 


8.70219 


0.0015175 


660 


435600 


287496000 


25.69047 


8.70659 


0015152 


661 


436921 


288804781 


25.70992 


8.71098 


0.0015129 


662 


438244 


290117528 


25.72936 


8.71537 


0.0015106 


663 


439569 


291434247 


25.74879 


8.71976 


0.0015083 


664 


440896 


292754944 


25.76S20 


8.72414 


0.0015060 


665 


442225 


294079625 


25.78749 


8.72852 


0.0015038 


666 


443556 


295408296 


25.80698 


8.73289 


0.0015015 


667 


444889 


296740963 


25.82634 


8.73726 


0.0014993 


668 


446224 


298077632 


25.84570 


8.74162 


0.0014970 


669 


447561 


299418309 


25.86503 


8.74598 


0.0014948 


670 


448900 


300763000 


25.8S436 


8.75034 


0.0014925 


671 


450241 


302111711 


25.90367 


8.75469 


0.0014903 


672 


451584 


303464448 


25.92296 


8.75904 


0.0014881 


673 


452929 


304821217 


25.94224 


8.76338 


0.0014859 


674 


454276 


306182024 


25.96151 


8.76772 


0.0014837 


675 


455625 


307546875 


25.98076 


8.77205 


0.0014815 


676 


456976 


308915776 


26 


8.77638 


0.0014793 


677 


458329 


310288733 


26.01922 


8.78071 


0.0014771 


678 


459684 


311665752 


26.03843 


8.78503 


0.0014749 


679 


461041 


313046839 


26.05763 


8.78935 


0.0014728 


680 


462400 


314432000 


26.07681 


8.79366 


0.0014706 


681 


463761 


315821241 


26.09598 


8.79797 


0.0014684 


682 


465124 


317214568 


26.11513 


8.80227 


0.0014663 


683 


466489 


318611987 


26.13427 


8.80657 


0.0014641 


684 


467856 


320013504 


26.15339 


8.81087 


0.0014620 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



55 



n 


n 2 


n 3 


V n 


3 

s/ n 


1 

n 


685 


469225 


321419125 


26.17250 


8.81516 


0.0014599 


686 


470596 


322828856 


26.19160 


8.81945 


0.0014577 


687 


471969 


324242703 


26.21068 


8.82373 


0.0014556 


688 


473344 


325660672 


26.22975 


8.82801 


0.0014535 


689 


474721 


327082769 


26.24881 


8.83229 


0.0014514 


690 


476100 


328509000 


26.26785 


8.83656 


0.0014493 


691 


477481 


329939371 


26.28688 


8.84082 


0.0014472 


692 


478864 


331373888 


26.30589 


8.84509 


0.0014451 


693 


480249 


332812557 


26.32489 


8.84934 


0.0014430 


694 


481636 


334255384 


26.34388 


8.85360 


0.0014409 


695 


483025 


335702375 


26.36285 


8.85785 


0.0014388 


696 


484416 


337153536 


26.38181 


8.86210 


0.0014368 


697 


485809 


338608873 


26.40076 


8.86634 


0.0014347 


698 


487204 


340068392 


26.41969 


8.87058 


0.0014327 


699 


488601 


341532099 


26.43861 


8.87481 


0.0014306 


700 


490000 


343000000 


26.45751 


8.87904 


0.0014286 


701 


491401 


344472101 


26.47640 


8.88327 


0.0014265 


702 


492804 


345948408 


26.49528 


8.88749 


0.0014245 


703 


494209 


347428927 


26.51415 


8.89171 


0.0014225 


704 


495616 


348913664 


26.53300 


8.89592 


0.0014205 


705 


497025 


350402625 


26.55184 


8.90013 


0.0014184 


706 


498436 


351895816 


26.57066 


8.90434 


0.0014164 


707 


499849 


353393243 


26.58947 


8.90854 


0.0014144 


708 


501264 


354894912 


26.60817 


8.91274 


0.0014124 


709 


502681 


356400829 


26.62705 


8.91693 


0.0014104 


710 


504100 


357911000 


26.64583 


8.92112 


0.0014085 


711 


505521 


359425431 


26.66458 


8.92531 


0.0014065 


712 


506944 


360944128 


26.68333 


8.92949 


0.0014045 


713 


508369 


362467097 


26.70206 


8.93367 


0.0014025 


714 


509796 


363994344 


26.72078 


8.93784 


0.0014006 


715 


511225 


365525875 


26.73948 


8.94201 


0.0013986 


716 


512656 


367061696 


26.75818 


8.94618 


0.0013966 


717 


514089 


368601813 


26.77686 


8.95034 


0.0013947 


718 


515524 


370146232 


26.79552 


8.95450 


0.0013928 


719 


516961 


371694959 


26.81418 


8.95866 


0.0013908 


720 


518400 


373248000 


26.83282 


8.96281 


0.0013889 


721 


519841 


374805361 


26.85144 


8.96696 


0.0013870 


722 


521284 


37(5367048 


26.87006 


8.97110 


0.0013850 


723 


522729 


377933067 


26.88866 


8.97524 


0.0013831 


724 


524176 


379503424 


26.90725 


8.97938 


0.0013812 



s« 



SQUARES, CtTBE§, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


V n 


3 

V n 


1 

n 


725 


525625 


381078125 


26.92582 


8.98351 


0.0013793 


726 


527076 


382657176 


26.94439 


8.98764 


0.0013774 


727 


528529 


384240583 


26.96294 


8.99176 


0.0013755 


728 


529984 


385828352 


26.98148 


8.99589 


0.0013736 


729 


531441 


387420489 


27 


9 


0.0013717 


730 


532900 


38901 7000 


27.01851 


9.00411 


0.0013699 


731 


534361 


390617891 


27.03701 


9.00822 


0.0013680 


732 


535824 


392223168 


27.05550 


9.01233 


0.0013661 


733 


537289 


393832837 


27.07397 


9.01643 


0.0013643 


734 


538756 


395446904 


27.09243 


9.02053 


0.0013624 


735 


540225 


397065375 


27.11088 


9.02462 


0.0013605 


736 


541696 


398688256 


27.12932 


9.02871 


0.0013587 


737 


543169 


400315553 


27.14771 


9.03280 


0.0013569 


738 


544644 


401947272 


27.16616 


9.03689 


0.0013550 


739 


546121 


403583419 


27.18455 


9.04097 


0.0013532 


740 


547600 


405224000 


27.20291 


9.04504 


0.0013514 


741 


549081 


406869021 


27.22132 


9.04911 


0.0013495 


742 


550564 


408518488 


27.23968 


9.05318 


0.0013477 


743 


552049 


410172407 


27.25803 


9.05725 


0.0013459 


744 


553536 


411830784 


27.27636 


9.06131 


0.0013441 


.745 


555025 


413493625 


27.29469 


9.06537 


0.0013423 


746 


556516 


415160936 


27.31300 


9.06942 


0.0013405 


747 


558009 


416832723 


27.33130 


9.07347 


0.0013387 


748 


559504 


418508992 


27.34959 


9.07752 


0.0013369 


749 


561001 


420189749 


27.36786 


9.08156 


0.0013351 


750 


562500 


421875000 


27.38613 


9.08560 


0.0013333 


751 


564001 


423564751 


27.40438 


9.08964 


0.0013316 


752 


565504 


425259008 


27.42262 


9.09367 


0.0013298 


753 


567009 


426957777 


27.44085 


9.09770 


0.0013280 


754 


568516 


428661064 


27.45906 


9.10173 


0.0013263 


755 


570025 


430368875 


27.47726 


9.10575 


0.0013245 


756 


571536 


432081216 


27.49545 


9.10977 


0.0013228 


757 


573049 


433798093 


27.51363 


9.11378 


0.0013210 


758 


574564 


435519512 


27.53180 


9.11779 


0.0013193 


759 


576081 


437245479 


27.54995 


9.12180 


0.0013175 


760 


577600 


43S976000 


27.56810 


9.12581 


0.0013158 


761 


579121 


440711081 


27.58623 


9.12981 


0.0013141 


762 


580644 


442450728 


27.60435 


9.13380 


0.0013123 


763 


582169 


444194947 


27.62245 


9.13780 


0.0013106 


764 


583696 


445943744 


27.64055 


9.14179 


0.0013089 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



57 



n 


n* 


n 3 


V n 


3 

V n 


1 

n 


765 


585225 


447697125 


27.65863 


9.14577 


0.0013072 


766 


586756 


449455096 


27.67671 


9.14976 


0.0013055 


767 


588289 


451217663 


27.69476 


9.15374 


0.0013038 


768 


589824 


452984832 


27.71281 


9.15771 


0.0013021 


769 


591361 


454756609 


27.73085 


9.16169 


0.0013004 


770 


592900 


456533000 


27.74887 


9.16566 


0.0012987 


771 


594441 


458314011 


27.76689 


9.16962 


0.0012970 


772 


595984 


460099648 


27.78489 


9.17359 


0.0012953 


773 


597529 


461889917 


27.80288 


9.17754 


0.0012937 


774 


599076 


463684824 


27.82086 


9.18150 


0.0012920 


775 


600625 


465484375 


27.83882 


9.18545 


0.0012903 


776 


602176 


467288576 


27.85678 


9.18940 


0.0012887 


777 


603729 


469097433 


27.87472 


9.19335 


0.0012870 


778 


605284 


470910952 


27.89265 


9.19729 


0.0012853 


779 


606841 


472729139 


27.91057 


9.20123 


0.0012837 


780 


608400 


474552000 


27.92848 


9.20516 


0.0012821 


781 


609961 


476379541 


27.94638 


9.20910 


0.0012804 


782 


611524 


478211768 


27.96426 


9.21303 


0.0012788 


783 


613089 


480048687 


27.98214 


9.21695 


0.0012771 


784 


614656 


481890304 


28 


9.22087 


0.0012755 


785 


616225 


483736625 


28.01785 


9.22479 


0.0012739 


786 


617796 


485587656 


28.03569 


9.22871 


0.0012723 


787 


619369 


487443403 


28.05352 


9.23262 


0.0012706 


788 


620944 


489303872 


28.07134 


9.22653 


0.0012690 


789 


622521 


491169069 


28.08914 


9.24043 


0.0012674 


790 


624100 


493039000 


28.10694 


9.24434 


0.0012658 


791 


625681 


494913671 


28.12472 


9.24823 


0.0012642 


792 


627264 


496793088 


28.14249 


9.25213 


0.0012629 


793 


628849 


498677257 


28.16026 


9.25602 


0.0012610 


794 


630436 


500566184 


28.17801 


9.25991 


0.0012594 


795 


632025 


502459875 


28.19574 


9.26380 


0.0012579 


796 


633616 


504358336 


28.21347 


9.26768 


0.0012563 


797 


635209 


506261573 


28.23119 


9.27156 


0.0012547 


798 


636804 


508169592 


28.24889 


9.27544 


0.0012531 


799 


638401 


510082399 


28.26659 


9.27931 


0.0012516 


800 


640000 


512000000 


28.28427 


9.28318 


0.0012500 


801 


641601 


513922401 


28.30194 


9.28704 


0.0012484 


802 


643204 


515849608 


28.31960 


9.29091 


0.0012469 


803 


644809 


517781627 


28.33725 


9.29477 


0.0012453 


804 


646416 


519718464 


28.35489 


9.29862 


0.0012438 



5§ 



SQUARES, CUBES. ROOTS, AND RECIPROCALS. 



n 


n 2 


n s 


*/~n 


3 

V n 


1 

n 


805 


648025 


521660125 


28.37252 


9.30248 


0.0012422 


806 


649636 


523606616 


28.39014 


9.30638 


0.0012407 


807 


651249 


525557943 


28.40775 


9.31018 


0.0012392 


808 


652864 


527514112 


28.42534 


9.31402 


0.0012376 


809 


654481 


529475129 


28.44293 


9.31786 


0.0012361 


810 


656100 


531441000 


28.46050 


9.32170 


0.0012346 


811 


657721 


533411731 


28.47806 


9.32553 


0.0012330 


812 


659344 


535387328 


28.49561 


9.32936 


0.0012315 


813 


660969 


537367797 


28.51315 


9.33319 


0.0012300 


814 


662596 


539353144 


28.53069 


9.33702 


0.0012285 


815 


664225 


541343375 


28.54820 


9.34084 


0.0012270 


816 


665856 


543338496 


28.56571 


9.34466 


0.0012255 


817 


667489 


545338513 


28.58321 


9.34847 


0.0012240 


818 


669124 


547343432 


28.60070 


9.35229 


0.0012225 


819 


670761 


549353259 


28.61818 


9.35610 


0.0012210 


820 


672400 


551368000 


28.63564 


9.35990 


0.0012195 


821 


674041 


553387661 


28.65310 


9.36270 


0.0012180 


822 


675684 


555412248 


28.67054 


9.36751 


0.0012165 


823 


677329 


557441767 


28.68798 


9.37130 


0.0012151 


824 


678976 


559476224 


28.70540 


9.37510 


0.0012136 


825 


680625 


561515625 


28.72281 


9.37889 


0.0012121 


826 


682276 


563559976 


28.74022 


9.38268 


0.0012107 


827 


683929 


565609283 


28.75761 


9.38646 


0.0012092 


828 


685584 


567663552 


28.77499 


9.39024 


0.0012077 


829 


687241 


569722789 


28.79236 


9.39402 


0.0012063 


830 


688900 


571787000 


28.80972 


9.39780 


0.0012048 


831 


690561 


573856191 


28.82707 


9.40157 


0.0012034 


832 


692224 


575930368 


28.84441 


9.40534 


0.0012019 


833 


693889 


578009537 


28.86174 


9.40911 


0.0012005 


834 


695556 


580093704 


28.87906 


9.41287 


0.0011990 


835 


697225 


582182875 


28.89637 


9.41663 


0.0011976 


836 


698896 


584277056 


28.91366 


9.42039 


0.0011962 


837 


700569 


586376253 


28.93095 


9.42414 


0.0011947 


838 


702244 


588480472 


28.94823 


9.42789 


0.0011933 


839 


703921 


590589719 


28.96550 


9.43164 


0.0011919 


840 


705600 


592704000 


28.98275 


9.43538 


0.0011905 


841 


707281 


594823321 


29 


9.43913 


0.0011891 


842 


708964 


596947688 


29.01724 


9.44287 


0.0011876 


843 


710649 


599077107 


29.03446 


9.44661 


0.0011862 


844 


712336 


6012115S4 


29.05168 


9.45034 


0.0011848 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



59 



n 


n 2 


n 3 


V n 


3 

V n 


1 

n 


845 


714025 


603351125 


29.06888 


9.45407 


0.0011834 


846 


715716 


605495736 


29.08608 


9.45780 


0.0011820 


847 


717409 


607645423 


29.10326 


9.46152 


0.0011806 


848 


719104 


609800192 


29.12044 


9.46525 


0.0011792 


849 


720801 


611960049 


29.13760 


9.46897 


0.0011779 


850 


722500 


614125000 


29.15476 


9.47268 


0.0011765 


851 


724201 


616295051 


29.17190 


9.47640 


0.0011751 


852 


725904 


618470208 


29.18904 


9.48011 


0.0011737 


853 


727609 


620650477 


29.20616 


9.48381 


0.0011723 


854 


729316 


622835864 


29.22328 


9.48752 


0.0011710 


855 


731025 


625026375 


29.24038 


9.49122 


0.0011696 


850 


732736 


627222016 


29.25748 


9.49492 


0.0011682 


857 


734449 


629422793 


29.27456 


9.49861 


0.0011669 


858 


736164 


631628712 


29.29164 


9.50231 


0.0011655 


859 


737881 


633839779 


29.30870 


9.50600 


0.0011641 


860 


739600 


636056000 


29.32576 


9.50969 


0.0011628 


86 L 


741321 


638277381 


29.34280 


9.51337 


0.0011614 


862 


743044 


640503928 


29.35984 


9.51705 


0.0011601 


863 


744769 


642735647 


29.37686 


9.52073 


0.0011587 


864 


746496 


644972544 


29.39388 


9.52441 


0.0011574 


865 


748225 


647214625 


29.41088 


9.52808 


0.0011561 


866 


749956 


649461896 


29.42788 


9.53175 


0.0011547 


867 


751689 


651714363 


29.44486 


9.53542 


0.0011534 


ms 


753424 


653972032 


29.46184 


9.53908 


0.0011521 


869 


755161 


656234909 


29.47881 


9.54274 


0.0011507 


870 


756900 


658503000 


29.49576 


9.54640 


0.0011494 


871 


758641 


660776311 


29.51271 


9.55006 


0.0011481 


872 


760384 


663054848 


29.52965 


9.55371 


0.0011468 


873 


762129 


665338617 


29.54657 


9.55736 


0.0011455 


874 


763876 


667627624 


29.56349 


9.56101 


0.0011442 


875 


765625 


669921875 


29.58040 


9.56466 


0.0011429 


876 


767376 


672221376 


29.59730 


9.56830 


0.0011416 


877 


769129 


674526133 


29.61419 


9.57194 


0.0011403 


878 


770884 


676836152 


29.63106 


9.57557 


0.0011390 


879 


772641 


679151439 


29.64793 


9.57921 


0.0011377 


880 


774400 


681472000 


29.66479 


9.58284 


0.0011364 


881 


776161 


683797841 


29.68164 


9.58647 


0.0011351 


882 


777924 


686128968 


29.69848 


9.59009 


0.0011338 


883 


779689 


688465387 


29.71532 


9.59372 


0.0011325 


884 


781456 


690807104 


29.73214 


9.59734 


0.0011312 



6o 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


V n 


3 ,— 

v n 


1 

n 


885 


783225 


693154125 


29.74895 


9.60095 


0.0011299 


886 


784996 


695506456 


29.76575 


9.60457 


0.0011287 


887 


786769 


697864103 


29.78255 


9.60818 


0.0011274 


S8S 


788544 


700227072 


29.79933 


9.61179 


0.0011261 


8S9 


790321 


702595369 


29.81610 


9.61540 


0.0011249 


890 


792100 


704969000 


29.83287 


9.619 


0.0011236 


891 


793881 


707347971 


29.84962 


9.62260 


0.0011223 


892 


795664 


709732288 


29.86637 


9.62620 


0.0011211 


893 , 


797449 


712121957 


29.88311 


9.62980 


0.0011198 


894 


799236 


714516984 


29.89983 


9.63339 


0.0011186 


895 


801025 


716917375 


29.91655 


9.63698 


0.0011173 


896 


802816 


719323136 


29.93326 


9.64057 


0.0011161 


897 


804609 


721734273 


29.94996 


9.64415 


0.001114S 


898 


806404 


724150792 


29.96665 


9.64774 


0.0011136 


899 


808201 


726572699 


29.98333 


9.65132 


0.0011123 


900 


810000 


729000000 


30 


9.65489 


0.0011111 


901 


811801 


731432701 


30.01666 


9.65847 


0.0011099 


902 


813604 


733870808 


30.03331 


9.66204 


0.0011086 


903 


815409 


736314327 


30.04996 


9.66561 


0.0011074 


904 


817216 


738763264 


30.06659 


9.66918 


0.0011062 


905 


819025 


741217625 


30.08322 


9.67274 


0.0011050 


906 


820836 


743677416 


30.09983 


9.67630 


0.0011038 


907 


822649 


746142643 


30.11644 


9.67986 


0.0011025 


908 


824464 


748613312 


30.13304 


9.68342 


0.0011013 


909 


826281 


751089429 


30.14963 


9.68697 


0.0011001 


910 


828100 


753571000 


30.16621 


9.69052 


0.00109S9 


911 


829921 


756058031 


30.18278 


9.69407 


0.0010977 


912 


831744 


758550528 


30.19934 


9.69762 


0.0010965 


913 


833569 


761048497 


30.215S9 


9.70116 


0.0010953 


914 


835396 


763551944 


30.23243 


9.70470 


0.0010941 


915 


837225 


766060875 


30.24897 


9.70824 


0.0010929 


916 


839056 


768575296 


30.26549 


9.71177 


0.0010917 


917 


840889 


771095213 


30.28201 


9.71531 


0.0010905 


918 


842714 


773620632 


30.29851 


9.71884 


0.0010893 


919 


844561 


776151559 


30.31501 


9.72236 


0.0010881 


920 


846100 


778688000 


30.33150 


9.72589 


0.0010870 


921 


848241 


781229961 


30.34798 


9.72941 


0.0010858 


922 


850084 


783777448 


30.36445 


9.73293 


0.0010846 


923 


851929 


786330467 


30.38092 


9. 73645 


0.0010834 


924 


S53776 


788889024 


30.39737 


9.73996 


0.0010S23 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



61 



n 


n 2 


n 3 


*/n~ 


V n 


1 

n 


925 


855025 


791453125 


30.41381 


9.74348 


0.0010811 


920 


857470 


794022770 


30.43025 


9.74099 


0.0010799 


927 


859329 


790597983 


30.44007 


9.75049 


0.0010787 


928 


801184 


799178752 


30.40309 


9.75400 


0.0010776 


929 


803041 


801705089 


30.47950 


9.75750 


0.0010764 


930 


804900 


804357000 


30.49590 


9.70100 


0.0010753 


931 


800701 


800954491 


30.51229 


9.70450 


0.0010741 


932 


808024 


809557508 


30.52808 


9.70799 


0.0010730 


933 


870489 


812100237 


30.54505 


9.77148 


0.0010718 


934 


872350 


814780504 


30.50141 


9.77497 


0.0010707 


935 


874225 


817400375 


30.57777 


9.77840 


0.0010095 


930 


870090 


820025856 


30.59412 


9.78295 


0.0010084 


937 


877909 


822650953 


30.01040 


9.78543 


0.0010672 


938 


879844 


825293072 


30.02079 


9.78891 


0.0010661 


939 


881721 


827930019 


30.04311 


9.79239 


0.0010050 


940 


883000 


830584000 


30.05942 


9.79580 


0010038 


941 


885481 


833237021 


30.07572 


9.79933 


0.0010027 


942 


887304 


835890888 


30.09202 


9.80280 


0.0010010 


943 


889249 


838501807 


30.70831 


9.80027 


0.0010004 


944 


891130 


841232384 


30.72458 


9.80974 


0.0010593 


945 


893025 


843908025 


30.74085 


9.81320 


0.0010582 


940 


894910 


840590530 


30.75711 


9.81000 


0.0010571 


947 


890809 


849278123 


30.77337 


9.82012 


0.0010500 


948 


898704 


851971392 


30.78901 


9.82357 


0.0010549 


949 


900001 


854070349 


30.80584 


9.82703 


0.0010537 


950 


902500 


857375000 


30.82207 


9.83048 


0.0010520 


951 


904401 


800085351 


30.83829 


9.83392 


0.0010515 


952 


900304 


802801408 


30.85450 


9.83737 


0.0010504 


953 


908209 


805523177 


30.87070 


9.84081 


0.0010493 


954 


910110 


808250664 


30.88089 


9.84425 


0.0010482 


955 


912025 


870983875 


30.90307 


9.84709 


0.0010471 


950 


913930 


873722816 


30.91925 


9.85113 


0.0010400 


957 


915849 


876467493 


30.93542 


9.85456 


0.0010449 


958 


917704 


879217912 


30.95158 


9.85799 


0.0010438 


959 


919081 


881974079 


30.90773 


9.86142 


0.0010428 


900 


921000 


887430000 


30.98387 


9.86485 


0.0010417 


901 


923521 


887503081 


31 


9.86827 


0.0010400 


902 


925444 


890277128 


31.01012 


9.87169 


0.0010395 


903 


927309 


893050347 


31.03224 


9.87511 


0.0010384 


904 


929290 


895841344 


31.04835 


9.87853 


0.0010373 



62 



SQUARES, CUBES, ROOTS, AND RECIPROCALS. 



n 


n 2 


n 3 


*/~n 


3 


n 


965 


931225 


898632125 


31.06445 


9.88195 


0.0010363 


966 


933156 


901428696 


31.08054 


9.88536 


0.0010352 


967 


935089 


904231063 


31.09662 


9.88877 


0.0010341 


968 


937024 


907039232 


31.11270 


9.89217 


0.0010331 


969 


938961 


909853209 


31.12876 


9.89558 


0.0010320 


970 


940900 


912673000 


31.14482 


9.89898 


0.0010309 


971 


942841 


915498611 


31.16087 


9.90238 


0.0010299 


972 


944788 


918330048 


31.17691 


9.90578 


0.0010288 


973 


946729 


921167317 


31.19295 


9.90918 


0.0010277 


974 


948676 


924010424 


31.20897 


9.91257 


0.0010267 


975 


950625 


926859375 


31.22499 


9.91596 


0.0010256 


976 


952576 


929714176 


31.24100 


9.91935 


0.0010246 


977 


954529 


932574833 


31.25700 


9.92274 


0.0010235 


978 


956484 


935441352 


31.27299 


9.92612 


0.0010225 


979 


958441 


938313739 


31.28898 


9.92950 


0.0010215 


980 


960400 


941192000 


31.30495 


9.93288 


0.0010204 


981 


962361 


944076141 


31.32092 


9.93626 


0.0010194 


982 


964324 


946966168 


31.33688 


9.93964 


0.0010183 


983 


966289 


949862087 


31.35283 


9.94301 


0.0010173 


984 


968256 


952763904 


31.36877 


9.94638 


0.0010163 


985 


970225 


955671625 


31.38471 


9.94975 


0.0010152 


986 


972196 


958585256 


31.40064 


9.95311 


0.0010142 


987 


974169 


961504803 


31.41656 


9.95648 


0.0010132 


988 


976144 


964430272 


31.43247 


9.95984 


0.0010121 


989 


978121 


967361669 


31.44837 


9.96320 


0.0010111 


990 


980100 


970299000 


31.46427 


9.96655 


0.0010101 


991 


982081 


973242271 


31.48015 


9.96991 


0.0010091 


992 


984064 


976191488 


31.49603 


9.97326 


0.0010081 


993 


986049 


979146657 


31.51190 


9.97661 


0.0010070 


994 


988036 


982107784 


31.52777 


9.97996 


0.0010060 


995 


990025 


985074875 


31.54362 


9.98331 


0.0010050 


996 


992016 


988047936 


31.55947 


9.98665 


0.0010040 


997 


994009 


991026973 


31.57531 


9.98999 


0.0010030 


998 


996004 


994011992 


31.59114 


9.99333 


0.0010020 


999 


998001 


997002999 


31.60696 


9.99667 


0.0010010 



Botes on Hlcjebra. 



Algebra is that branch of mathematics in which the quan- 
tities are denoted by letters and the operations to be performed 
upon them are indicated by signs. The same signs are used to 
indicate the same operations as in arithmetic. 

Signs of Quantity and Signs of Operation. 

If a quantity is written a + ( — b), the sign that precedes the 
parenthesis is called the sign of operation, and the sign within 
the parenthesis is called the sign of quantity with respect to b, 
but expressions of this kind can be reduced to have only one 
sign. Thus, a + ( — b) = a — b and this final sign is called the 
essential sign. 

a + (-F b) = a -f b. 

a + ( — b) = a — b. 
a — ('+ b) = a — b. 
a — ( — b) = a -|- b. 

Thus, when the sign of operation and the sign of quantity 
are alike the essential sign is +, but if they are unlike the es- 
sential sign is — . 

In multiplying any two quantities, like signs in the two 
factors give -f- in the product, but unlike signs in the two factors 
give — in the product; thus, (-f- a) X ( — b) = — ab, and ( — a) 
X (— b) = ab. 

In division, like signs in dividend and divisor give + in the 
quotient, but unlike signs in dividend and divisor give — in the 
quotient; thus: 

— a a 



= +4- 



— b b 

Useful Formulas and Rules in Algebra. 

The following rules are very useful to remember in solv- 
ing practical problems in algebra. Let a and b represent any 
two quantities ; then a + b will represent their sum and a — b 
their difference ; then (a + b) X (a + b) = a 2 + 2 ab + b 2 . 

{a + b) X (a + b) is also written {a + b) 2 . 
(6 3 ) 



64 NOTES ON ALGEBRA. 

This rule reads : 

The square of the sum of any two quantities is equal to 
the square of the first quantity plus double the product of both 
quantities, phis the square of the second quantity. 

{a — b) X (a —b) = (a — b) 2 = a 2 —2ab J r b 2 . 

This rule reads: 

The square of the difference of any two quantities is equal 
to the square of the first quantity minus twice the product of 
both quantities, plus the square of the second quantity. 

{a + b) X {a — b) = a 2 — b 2 . 

This rule reads : 

The sum of any two quantities multiplied by their differ- 
ence is equal to the difference of their squares. 

Extracting Roots. 

An even root of a positive quantity is either + or — . An 
even root cannot be extracted of a negative quantity, as \/ a 2 
may be either a or — a ; but V — a 2 is impossible, because 
(—'a) X (— a) = a 2 and (— a) X (+ a) = a 2 . 

An odd root may be extracted as well of a negative quantity 
as a positive quantity, and the sign of the root is always the 
same as the sign of the quantity before the root was extracted. 

1 3 

Thus : s/a z == a. but V ( — a) n = ( — a). 

Powers. 

When a number or a quantity is to be multiplied by itself 
a given number of times, the operation is indicated by a small 
number at the right-hand corner of the quantity : for instance, 
a' z = a X a. 

A quantity of this kind is called a power ; the small number 
is called the exponent, or the index of the power. Two powers 
of the same kind may be multiplied by adding the exponents ; 
for instance, a 2 a 3 = a 2rZ = a h . 

Two powers of the same kind may be divided by subtract- 
ing their exponents ; for instance, 

— = a 5 "- = a 3 = a X a X a. 

a- 



a* 
~a* 



-3 = a 2 = a X a. 



a* Z^, 1 

_ = a 73 —* = a 1 = a. 

a i 

JL_ ao-o = a - = 1. 
a° 



NOTES ON ALGEBRA. 65 

Thus, any quantity in power must be 1, because always 
when dividend and divisor are alike the quotient must be 1. 

a 5 

-— = a 5-6 = a -1 , but a 5 divided by a 5 is equal to 1 ; therefore 



<z 6 

4- must be J- ; * 
a 6 a a 1 a 2 



Thus, any quantity with a negative exponent is one divided 
by that quantity considering the exponent as positive. We 
may, therefore, say that as a positive exponent indicates how 
many times a quantity is to be used as a factor, a negative ex- 
ponent indicates how many times a quantity should be used as 
a divisor ; for instance, 

a-* = -L ; ar-a = 1 . ; or* = * 

a a X a a Xa X a 

Thus: 
6-1 = 1; 6-2 = (|)2 = ^ ; 3-2 = (1)2 = 1, etc. 

Equations. 

An algebraic expression of equality between two quantities 
is called an equation. The two quantities are connected by the 
sign of equality, and the quantity on the left-hand side of the 
sign is called the first member, and the quantity on the right- 
hand side is called the second member of the equation. 

If a part or all of the known quantities are expressed in 
letters it is called a literal equation: ax = a + b — 2 c is a 
literal equation. 

If the equation contains no letters except the unknown 
quantity, usually expressed by x, it is called a numerical equa- 
tion ; for instance, 5 x = 12 — 7 + 9, is a numerical equation. 

In solving equations, we may, without destroying the equal- 
ity of the equation, add an equal quantity to both members, 
subtract an equal quantity from both members, multiply both 
members by an equal quantity, divide both members by an 
equal quantity, extract the same root of both members, raise 
both members to the same power. 

Quantities inclosed by parentheses, a bar, or under a radical 
sign, and quantities connected by the sign of multiplication, 
must always be considered and operated upon as one quantity. 

Example 1. 

jc = 3X8 + 10- 3 + 3X10. 
x = 24 + 10 — 3 + 30. 

* = C4— 3. 

* = 61. 



-.- vizis IN" .-zizi?.-.. 

Example 2. 

b — * — :: 



I::., 



- : . 



..- = :: « 



( j 



- : 



; 



■' 



~» x (t- + ") 

= 12x(l-§- + «) 

= isx(*f-) 



-'..-. " LZ -= 



= :- 



- — : ; - :: 
- 



. - : 



z. ZjZI? 1Z 



: - if ) « 



— 1 x(s- 3) + VST 


— 1 : 


— 1 : - v . f » - ; 




j=|6X 5-S-6) X - 




X = -f 




= "- 





iiXA . : 



:.i' 



NOTES ON ALGEBRA. 67 

Call small gear x and large gear y ; then x 4- y must be 108, 
because 6|^ X 16 = 108, which is the number of teeth in both 
gears added together. The ratio is 1 to 3; therefore 3x=y. 

Thus : 

x + y = 108, transposed to 

x 4- 3 x = 108. 

4 *■ = 108. 

.r= Jos 
4 

jr = 27 teeth for small gear. 
The large gear = 27 X 3 = 81 teeth. 

Quadratic Equations. 

Equations containing one or more unknown quantities in 
the second power are called quadratic equations. If the un- 
known quantity only exists in the second power the equation 
may be brought to the form x' 2 = a and x = \Z^T 

This square root may be either plus or minus. 

If the unknown quantity exists in both the first and the 
second power the equation may be brought to the form x 2 -\-a x 
= by or it may be brought to the form x 2 — a x = b. 

The coefficient a may be any number. After the equation 
is brought to this form, complete the square of the first member 
by adding to both members of the equation the square of half 
the coefficient a ; this will make the left member of the equa- 
tion a complete square. 

Example. 

A coal bin is to hold six tons of coal. Allow 40 cubic 
feet per ton. (It takes 35 to 40 cubic feet to hold a ton of 
coal in a bin). Make the width of the bin 6 feet, and the 
length equal to the width and the depth added together. How 
deep and how long will the bin be? 

Depth = x and length =y. 

6 x y = 240, because 6 times 40 equals 240. 

6 4- x = y, because width + depth = length. 

Thus: 

6 x (6 + x) = 240. 

6 x 2 + 36 x = 240. 
Dividing by 6 we have : x 2 4- 6x = 40. 

Completing the square : x 2 + Qx + 3 2 = 40 + 3 2 . 

:r + 3= \/ 40 4- 9. 
Extracting the square root: x + 3 = \f 49. 

* + 3 = 7. 

:r = 7_3. 

x = 4 feet deep. 



68 NOTES ON ALGEBRA. 

Length = width + depth, =6+4, =10 feet. 
This satisfies the conditions of the problem, because 
C X 4 X 10 = 240 cubic feet, 
and the width and depth added together equal the length. 

Progressions. 

A progression is a series of numbers increasing or decreas- 
ing, according to a fixed law. 

The successive numbers of which the progression consists 
are called terms ; the first and the last terms are called the 
extremes and the others are called the means. 

ARITHMETICAL PROGRESSION. 

An arithmetical progression is a series of numbers which 
increase or decrease by a constant difference. For instance : 

2, 4, 6, 8, 10, 12, 14, 16, is an ascending series ; 
20, 18, 16, 14, 12, 10, 8, is a descending series. 

In each of these series the common difference is 2. 

The following are the elements considered in an arith- 
metical progression : 

a = First term ; / = last term ; d = the common difference ; 
n = the number of terms ; s = the sum of all the terms. 

When any three of these quantities are known the other two 
may be calculated : 

In the above example of an ascending series : 
a = 2; /=16; d=2; » = 8 ; s = 72. 



Formulas : 


Examples : 


a — l—(n — l)Xd 


a = 16 — (8 — 1) X 2 = 2 


l = a+ (a — 1) X d 


/ = 2 + (8 — 1) X 2 = 16 


d- l ~ a 
n — 1 


d~ 16 ~ 2 -2 
8 — 1 


/ — a i -, 
n — — — + 1 
d 


„_16-2 + 

2 


2 


, _ (2 + 16) X 8 _ 72 
2 



NOTES ON ALGEBRA. 69 

In the above example of a descending series : 

a = 20; 1=8] d=2-, n = *1; s = 9S. 



Formulas : 




E: 


samples : 


a = /+(« — 1) 


X d 




a = 8 + (7 — 1) X 2 = 20 


/= a — (n — 1) 


X d 




/= 20 — (7 — 1) X 2 = 8 


d- a ~ l 

11— \ 






d- 20 ~ 8 ~2 
7 — 1 


« = — j — + 1 

u 






„-20~8 +1=7 
2 


2 






2 



GEOMETRICAL PROGRESSION. 

A Geometrical Progression is a series of numbers which 
increase or decrease by a common constant ratio. For 
instance : 

3, 6, 12, 24, 48, is an ascending series ; 48, 24, 12, 6, 3 is a 
descending series. 

The following are the elements considered in a geometrical 
progression : 

a = first term ; / = last term ; r = ratio ; n = number 
of term ; s = sum of the terms. 

When three of these are known the other two may be cal- 
culated. 

In this example of an ascending series : 

a = 3 ; / = 48 ; r = 2 ; n = 5; s=9S. 
Formulas : Examples : 

/ _ 48 _ 48 _ 48 _ 3 

r n-i 2 5 ~ 1 2 4 16 

IzzzaXr*- 1 /=3 X 2 ^ = 3 X 16 =48 

n - 1 5-1 _ 4 

r = ^-J- r = ^-^- = ^Jl6 = 2 

„- log. 4S- log 3 + ± = 

«._. log.l-log.a M 1 ^- 2 

JogTr 1.681241 — 0.477121 1 x __ 5 

0.30103 



_ / X r — a _ 48 X 2 — 3 _ Q3 

r— 1 " 2—1 



70 NOTES ON ALGEBRA. 

In this example of a descending series : 

a = 48 ; / = 3 ; r = 2 ; n = 5 : J = 93. 

Formulas : Examples : 

a = / X r* -* a = 3 X 2 &-* = 3 X 16 = 48 

j_ a j_ 48_48_48_ 3 

r n~i 2 5 - 1 2* 16 

w _ /^. 48 — /qg. 3 _ 

/qg. a — /^. / log. 2 "+" 

/^T7 +1 1.681241 — 0.477121 _. t _ q 

0.30103 



n 



a X r — / 48X2 — 3 

r — 1 • y— 2 — l — yd 

The Arithmetical Mean. 

The arithmetical mean of two or more quantities is obtained 
by adding the quantities and dividing the sum by their number. 

For instance, the arithmetical mean of 14 and 16 is 14 ^~ 16 . — ^5 
Thus : The arithmetical mean is simply the average. 

The Geometrical Mean. 

The geometrical mean of two quantities is the square root 
of their product. For instance, the geometrical mean of 14 and 
16 is Vl4 X 16 = 14.9666. 

The geometrical mean of two numbers is also called their 
mean proportional. 

When the difference between two numbers is small as com- 
pared to either of them, their arithmetical mean is approxi- 
mately equal to their geometrical mean. 

This fact may be used to advantage for calculating approxi- 
mately a root of any number. 

For instance, find the square root of 148. 

Knowing that the square of 12 is 144, twelve is used as a 
divisor, thus: 

^ = 12.333, and 1 -M?|±i- 2 = 12.168, 

which is correct within 0.005. 



3Looaritbms. 



Logarithms are a series of numbers computed in order to 
facilitate all kinds of laborious calculations, such as evolution, 
involution, multiplication and division. 

Addition takes the place of multiplication, subtraction the 
place of division; multiplication that of involution, and divi- 
sion of evolution. 

The logarithm of any given number is the exponent of the 
power to which another fixed number, called the base, must be 
raised in order to produce the given number. 

There are two systems of logarithms in more or less general 
use in mechanical calculations : namely, the Napierian system 
and the Briggs system. 

The Napierian system of logarithms was invented and 
tables published by Baron John Napier, a Scotch mathema- 
tician, in 1614, but these tables were improved by John Speidell 
in 1619. 

The modulus of any system of logarithms is a constant by 
which the Napierian logarithm of any given number must be 
multiplied in order to obtain the logarithm for the same number 
in the other system. 

The base of the Napierian system of logarithms is an in- 
commensurable number expressed approximately by 2.718281828. 
In mathematical works this base is usually denoted by the 
letter e. 

The Napierian logarithms are frequently called hyperbolic 
logarithms, from their relation to certain areas included between 
the equilateral hyperbola and its asymptotes. 

The Napierian logarithms are sometimes called natural 
logarith?ns. 

The Briggs system of logarithms was first invented and 
computed by Professor Henry Briggs of London in 1615, and 
is usually termed the common system of logarithms. When- 
ever logarithms in general is mentioned the Briggs system is 
always the one referred to. 

The Briggs system of logarithms has for its modulus 
0.4342945, and 10 for its base. Therefore the Briggs logarithm 
of a number is the exponent of the power to which 10 must be 
raised in order to give the number. Thus : 

(7i) 



11 



Loj 



1=0 because 10° = 


1. 


10 = 1 " 10 1 = 


10. 


100 = 2 " 10 2 = 


100. 


,000 = 3 " 10 3 = 


1,000. 



10,000 



10 4 = 10,000. 



The logarithm of any number between 1 and 10 is a frac- 
tion smaller than 1. The logarithm of any number between 10 
and 100 is a number between 1 and 2. The logarithm of any 
number between 100 and 1000 is a number between 2 and 3, etc. 

The decimal part of the logarithms is called the mantissa, 
and is given in the table commencing on page 88. 

The integer part of a logarithm is called the index or some- 
times the characteristic, and is not given in the table, but 
is obtained by the rule that it is one less than the number 
of figures in the integer part of the number ; thus, the index of a 
logarithm for any number consisting of two figures must be 1; 
the index of the logarithm for a number consisting of three 
figures must be 2, etc. 

The index of the logarithm of a decimal fraction is a nega- 
tive number. Sometimes the negative index is denoted by 

writing a minus sign over it; for instance, log. 0.5240 = 1.719331, 
or the negative index is denoted by writing it after the mantissa ; 
thus, log. 0.5240 = 9.719331 — 10. This, of course, is of exactly 
the same value whether written — 1 or 9 — 10. Either of 
these expressions is minus one in value, but it is more con- 
venient in logarithmic calculations to write the negative index 

after the mantissa ; thus, instead of writing 1, write 9 

— 10 ; instead of 2, write 8 — 10, etc. Only the mantissa 

is given in the table, but the index (as already explained) is 
obtained by the rule: One less than the number of figures on 
the left side of the decimal fioitit . Therefore, in order to 
memorize and explain this rule, the following examples are 
inserted : 



Number. Logarithm. 


Number. 


Logarithm. 


8236 
823.6 
82.36 
8.236 
0.82-36 


3.915716 
2.915716 
1.915716 
0.915716 
9.915716 — 10 


0.08236 
0.008236 
. 0.0008236 
0.00008236 
0.000008236 


8.915716—10 
7.915716 — 10 
6.915716 — 10 
5.915716 — 10 
4.915716—10 



Multiplying or dividing a number by any power of 10 
does not change the mantissa in the corresponding logarithm, 
but only the index ; for instance : 



LOGARITHMS. 73 

Log. 0.5 = 9.698970 — 10, 
and Log. 500 = 2.698970, etc. 

Thus, the mantissa of a logarithm is the same whether the 
number is 0.5, 5, 50, 500, 5,000, etc. It is only the index that 
is changed ; therefore, when a number consists of three or less 
figures, its logarithm is found in the tables by taking the 
mantissa found in the first column to the right of the number ; 
that is, in the column under cipher. The index is found by the 
same rule as before. For instance, logarithm to 537 will be 
2.729974. 

To Find the Logarithm of a Number Consisting of 
Four Figures. 

First find the figures in the column headed " N " corre- 
sponding to the first three figures of the number ; in line with 
these figures, in the column headed by the fourth figure, will be 
found the mantissa of the logarithm corresponding to the 
complete number. By prefixing the index, according to the 
rule already given, the complete logarithm is obtained. 

Example. 

Find logarithm of 5375. 

Solution : 

Under the heading " N " find 537 ; and in the column at the 
top of the table find "5"; under 5 in the line with 537 is 
730378. 

This is the mantissa of the logarithm. The index for a 
number consisting of four integers is 3, therefore the complete 
logarithm of 5375 is 3.730378. 

To Find the Logarithm of a Number Having Hore Than 

Four Figures. 

Example 1. 

Find the logarithm to 3658.2. 

Solution : 

Log. 3658 = 3.563244 and log. 3659 = 3.563362 ; therefore the 
logarithm for 3658.2 must be somewhere between the two logar- 
ithms thus found in the table. The difference between these 
two logarithms is 0.000118; that is, if the number is increased 
by 1 the logarithm increases 0.000118, therefore if this number 
is increased 0.2 the corresponding logarithm must increase 0.2 
times, 0.000118 = 0.0000236, which may be taken as 0.000024. 

Thus : 

Log. 3658 = 3.563244 
Difference corresponding to 0.2 = 0.000024 

Log. 3658.2 = 3.563268 



74 LOGARITHMS. 

It is unnecessary to calculate the difference, as the average 
difference between the logarithms in each line is given in the 
column headed " D" in the tables. The difference in this case 
is given in the table as 119. 

Example 2. 

Find logarithm to 1892.5. 

Solution : 

The mantissa of the number 1892 is given in the table as 
276921. The difference is given as 229. The index for a num- 
ber consisting of four integers is 3. Thus : 

Log. 1892 =3.276921 
0.5 X 0.000229 = 0.0001145 = 115 

Log. 1892.5 = 3.277036 

Example 3. 

Find logarithm to 85673. 

Solution: 

The mantissa for the number 85670 is given in the table as 
932829. The difference is given in the table as 51. The index 
for a number consisting of five integers is 4. 

When an increase of 10 in the number increases the log- 
arithm 0.000051 an increase of 3 must increase the correspond- 
ing logarithm 0.3 times 0.000051. Thus: 

Log. 85670 = 4.932829 
0.3 X 0.000051 = 0.0000153 = 15 

Log. S5673 = 4.932844 

These calculations (or interpolations as they are usually 
called) are based upon the principle that the difference between 
the numbers and the difference between their corresponding 
logarithms are directly proportional to each other. This, how- 
ever, is not strictly true; but within limits, as it is used here, 
it is near enough for practical results. 

To Find the Number Corresponding to a Given 
Logarithm. 

Example 1. 

Find the number corresponding to the logarithm 2.610979, 

Solution : 

Always remember when looking for the number not to 
consider the index, but find the mantissa 610979 in the table. 
In the same line as this mantissa, under the heading " N," is 
408, and on the top of the table in the same column as this 
mantissa is 3 ; thus, the number corresponding to this mantissa 
is 4083 and the index of the logarithm is 2 ; consequentiy the 



LOGARITHMS. 75 

number is to have three figures on the left-hand side of the 
decimal point ; thus, the number corresponding to the logarithm 
2.610979 will be 408.3. 

Example 2. 

Find the number corresponding to the logarithm 3.883991. 

Solution : 

This mantissa is not in the table. The nearest smaller 
mantissa is 883945, and to this mantissa corresponds the number 
7655. The nearest larger mantissa is 884002, and to this 
corresponds the number 7656. 

Thus, an increment in the mantissa of 57 increases the 
number by 1, but the difference between the mantissa 883945 
and 883991 is 46, therefore the number must increase ff =0.807. 

Number of Log. 3.883945 = 7655 

Difference 0.000046 = 0.807 



Number of Log. 3.883991 = 7655.807 
Addition of Logarithms. 

(MULTIPLICATION.) 

Where the logarithms of the factors have positive indexes, 
add as if they were decimal fractions, and the sum is the log- 
arithm corresponding to the product. 
Example 1. 

Multiply 81 by 65 by means of logarithms. 
Solution : 

Log. 81 = 1.908485 
Log. 65 = 1.812913 

3.721398 
and to this mantissa corresponds the number 5265. The index 
is 3; therefore the number has no decimals, as it consists of only 
four figures. 

To Add Two Logarithms when One Has a Positive and 
the Other a Negative Index. 

Example 2. 

Multiply 0.58 by 32.6 by means of logarithms. 

Solution : 

Log. 0.58 = 9.763428 — 10 
Log. 32. 6 = 1.513218 

11.276646 — 10 



76 LOGARITHMS. 

This reduces to 1.276646 and to this logarithm corresponds 
the number 18.906. This mantissa. 276646. cannot be found in the 
table, but the nearest smaller mantissa is 276462. and the differ- 
ence between this and the next is found by subtraction to be 
230, and the difference between this and the given mantissa 
is 1S4. 

Thus: 

Given logarithm 1.276646 
To the tabulated log. 1.276462 corresponds 18.90 

Difference 0.000184 gives 0.008 

Thus, logarithm 1.276646 gives number 18.908 

To Add Two Logarithms, Both Having a Negative 

Index. 

Add both logarithms in the same manner as decimal frac- 
tions, and afterwards subtract 10 from the index on each side 
of the mantissa. 

Example. 

Multiply 0.S2 by 0.082 by means of logarithms. 

Solution : 

Log. 0.82 = 9.913814 — 10 

Log. 0.082 = 8.913814 — 10 

18.827628 — 20 

By subtracting 10 on each side of the mantissa this logar- 
ithm reduces to 8.827628 — 10 and to the mantissa 827628 corre- 
sponds the number 6724. but the negative index S — 10 

indicates that this first figure 6 is not a whole number, but that 
it is six-hundredths : therefore a cipher must be placed between 
this 6 and the decimal point in order to give 6 the right value 
according to the index: thus, to the logarithm 8.827628 — 10 
corresponds the number 0.06724. 

Subtraction of Logarithms. 

(DIVISION.) 

Logarithms are subtracted as common decimal fractions. 

To Subtract Two Logarithms, Both Having a Positive 

Index. 

Example. 

Divide 490 by 70 by means of logarithms. 

Solution : 

Log. 490 = 2.690196 
Log. 70 — 1.845098 

0.845098 



LOGARITHMS. 77 

and to the mantissa of this logarithm corresponds the number 
7 or 70 or 700 or 7000, etc., in the table of logarithms, but the 
index of this logarithm is a cipher; therefore the answer must 
be a number consisting of one figure, thus it must be 7. 

To Subtract a Larger Logarithm From a Smaller One. 

This is the same as to divide a smaller number by a larger 
one. Before the subtraction is commenced add 10 to the index 
of the smaller logarithm (that is, to the minuend) and place 
— 10 after the mantissa, then proceed with the subtraction as 
if they were decimal fractions. 

Example. 

Divide 242 by 367 by means of logarithms. 

Solution : 

Log. 242 = 2.383815 = 12.383815 — 10 
Log. 367 = 2.564666 

9.819149 — 10 

and to the mantissa of this logarithm corresponds, according 
to the table, the number 6594, but the negative index, 9 — 10, 
indicates it to be 0.6594. 

Thus, 242 divided by 367 = 0.6594. 

Multiplication of Logarithms. 

(INVOLUTION.) 

To multiply a logarithm is the same as to raise its corre- 
sponding number into the power of the multiplier. 

Logarithms having a positive index are multiplied the same 
as decimal fractions. Thus : 

Square 224 by means of logarithms. 
Solution: 

2 X log. 224 = 2 X 2.350248 = 4.700496 = 50176 

Logarithms having a negative index are multiplied the 
same as decimal fractions, but an equal number is subtracted 
from both the positive and the negative parts of the logarithm, 
in order to bring the negative part of the index to — 10. 

Example 1. 

Square 0.82 by means of logarithms. 

Solution: 

2 X log. 0.82 = 2 X (9.913814 — 10) = 19.827628 — 10, and 
subtracting 10 from both the positive and the negative parts of 
the logarithm, the result is 9.827628 — 10 ; this gives the num- 
ber 0.6724. 



7 8 logarithms. 

Example 2. 

Raise 0.9 to the 1.41 power. 
Solution : 

1.41 X log. 0.9 = 1.41 X (9.954243 — 10) == 14.035483 — 14.1 

In this_ example 10 cannot be subtracted from both parts 
of the logarithm, but 4.1 must be subtracted in order to get — 10, 
after the subtraction is performed. The logarithm will then 
read 9.935483 — 10, which corresponds to the number 86195, and 
the negative index, 9 — 10, makes this 0.86195. 

Division of Logarithms. 

(EVOLUTION.) 

To divide a logarithm is the same as to extract a root of 
the number corresponding to the logarithm. 

Logarithms having a positive index are divided the same 
as common decimal fractions. 

Example. 

Extract the cube root of 512 by means of logarithms. 
Solution : 

l °Z' 512 = *??«? = 0.90309 

3 3 

and the number corresponding to this logarithm is 8, 80, 800, 
8,000, etc., but the index of this logarithm is a cipher ;. there- 
fore the answer must be a number consisting of one integer, 
consequently it must be 8. 

To Divide a Logarithm Having a Negative Index. 

Select and add such a number to the index as will give 
10 without a remainder for the quotient in the negative index 
on the right-hand side of the mantissa after division is per- 
formed. 

Example 1. 

Extract the square root of 0.64 by means of logarithms. 

Solution : 

log. 0.64 9.80618-10 19.80618-20 = 
2 ~ 2 2 

and to this logarithm corresponds the number 0.8. 

Example 2. 

Extract the cube root of 0,125 by means of logarithms. 



LOGARITHMS. 7 9 



Solution 



log. 0.125 _ 9.09691—10 _ 29.09691—30 _ 9 69g9>7 _ 1Q 
3 " 3 3 

and to this logarithm corresponds the number 0.5. 

Example 3. 

Extract the 1.7 root of 0.78. 

Solution : 

log. 0.78 9.892095 — 10 

177 ~ L7 

We cannot here, as in previous examples, add a multiple of 
10 to the index on each side of the mantissa, but 7 must be 
added in order that the negative quotient shall be — 10 after 
the division is performed. Thus: 

9.892095 — 10 16.892095 — 17 aQ «-<>« m 
— = 9.93oo2b — 10 

1.7 1.7 

and to this logarithm corresponds the number 0.864. 



Short Rules for Figuring by Logarithms. 

MULTIPLICATION. 

Add the logarithms of the factors and the sum is the logar- 
ithm of the product. 

DIVISION. 

Subtract divisor's logarithm from the logarithm of the divi- 
dend and the difference is the logarithm of the quotient. 

INVOLUTION. 

Multiply the logarithm of the root by the exponent of the 
power and the product is the logarithm of the power. 

Example. 
Log. 86 2 = 2 X log. 86 = 2 X 1.934498 = 3.868996 
and to this logarithm corresponds the number 7396. 

EVOLUTION. 

The logarithm of the number or quantity under the radical 
sign is divided by the index of the root, and the quotient is the 
logarithm of the root. 



8o LOGARITHMS. 

Example. 

* log. 2401 3.380392 

Log. -y 2401 = — =~i = j = 0.845098 

and this logarithm corresponds to the number 7. 

EXPONENTS. 

The logarithm of a power divided by the logarithm of the 
root is equal to the exponent of the power. 

Example. 

8* = 64 

log. 64 



x = 



X = 



log. 8 

1.80618 



0.90309 

x = 2 

The logarithm of a quantity under the radical sign divided 
by the logarithm of the root is equal to the index of the root. 

Example. x 

8 = \ZBl2 



x 



log. 512 
log 8 
2.70927 



0.90309 
x = 3 



The reason for these last rules may be understood by re- 
ferring to the rules for Involution and Evolution ; for instance : 

86 2 = 7396, and this expressed by logarithms is : 

2 X log. 86 = log. 7396. 

Therefore: ^ 7396 = 2. 
log. 86 

FRACTIONS. 

The logarithm of a common fraction is found, either by first 
reducing the fraction to a decimal fraction, or by taking the 
logarithm of the numerator and the logarithm of the denominator 
and subtracting the logarithm of the denominator from the log- 
arithm of the numerator ; the difference is the logarithm of 
the fraction. 



LOGARITHMS. 8 1 



Example. 



Log. % = log. 8 — log. 4 

Log. 3 = 0.477121 = 10.477121 — 10 

Log. 4 = 0.602060 

Thus, log. U = 9.875061 — 10 
This is also the logarithm of the decimal fraction 0.75. 



RECIPROCALS. 

Subtract the logarithm of the number from log. 1, which is 
10.000000 — 10, and the difference is the logarithm of the reci- 
procal. 

Example. 

Find the reciprocal of 315. 



Solution 



Log. 1 = 10.000000 — 10 
Log. 315 = 2.498311 



Log. reciprocal of 315 = 7.501689 — 10 

To this logarithm corresponds the decimal fraction 
0.0031746, which is, therefore, the reciprocal of 315. 



Simple Interest by Logarithms. 

Add logarithm of principal, logarithm of rate of interest, 
and logarithm of number of years ; from this sum subtract log- 
arithm of 100. The difference is the logarithm of the interest. 

Example. 

Find the interest of #800 at 4% in 5 years. 

Solution : 

Log. 800 = 2.90309 

Log. 4 = 0.60206 

T - 0.69897 

Log. o = 

s 4.20412 

Log. 100 = 2.00000 
Log. interest = 2.20412 = $160 = Interest. 



82 LOGARITHMS. 

Compound Interest by Logarithms. 

When the interest, at the end of each period of time, is 
added to the principal the amount will increase at a constant 
rate ; and this rate will be the amount of one dollar invested 
for one period of the time. For instance : If the periods of time 
be one year each, then $30 in 3 years at 5 % compound interest 
will be : 

#30 X 1.05 = #31.50 at the end of first year. 

$31.50 X 1.05 = $33,075 at the end of second year. 

$33,075 X 1.05 = $34.73 at the end of third year. 

This calculation may be written : 

#30 X 1.05 X 1.05 X 1.05 = $34.73 

which also may be written 

X (1.05) 3 = $34.73. 



Thus, compound interest is a form of geometrical progres- 
sion, and may be calculated by the following formulas : 

a =p X r n 

Log. a — n X log. r + log. p 
a 

Log. P = log. a — n X log. r 

n _ log, a — log. j> 
log. r 

Log. r = kg-*- l°& i> 

p = Principal invested. 

n = The number of periods of time. 

a = The amount due after 7i periods of time. 

r = The amount of $1 invested one period of time. 

Note. — The quantity r is always obtained by the rule: 

Divide the rate of interest per period of time by 100, and 
add 1 to the quotient. 

Example. 

What is the amount of $816 invested 6 years at 4% com- 
pound interest? 



LOGARITHMS. 83 

Solution by formula : 

Log. a = n X log. r 4- log. ft 

Log. a = 6 X log. 1.04 + log. 816 

Log. a = Q X 0.017033+ 2.911690 

Log. a = 0.102198 4- 2.911690 

Log. a = 3.013888 

a = ^1032.49 = Amount. 

Example. 

If J$750 is invested at 3% compound interest, how many 
years will it take before the amount will be 

Solution by formula : 

n= log- a —log-P 
log.r 

_ log. 950 — log. 750 
. log. 1.03 

= 2.977724 — 2.875061 

0.012837 

Example. 

A principal of $3750 is to be invested so that by compound 
interest it will amount to $5000 in six years. Find rate of 
interest. 

Solution by formula : 

Log. r = log-"-loZ-l> 



= 8 years (nearly). 



Log. r = 



n 
3.698970 — 3.574031 



Log. r = 0.020823 

r = 1.0491 

Rate of interest = lOOr — 100 = 100 X 1.0491 — 100 = 4.91% ; 
or 5 % per year (very nearly). 

Discount or Rebate. 

When calculating discount or rebate, which is a deduction 
upon money paid before it is due, use formula: 

fi= — 

Log. ft = log. a — n X log. r 



84 LOGARITHMS. 

Example. 

A bill of $500 is due in 3 years. How much cash is it worth 
if 3% compound interest should be deducted. 

Log. p = log. a — n X log. r 
Log.p = 2.698970 — 3 X 0.012837 
Log. p = 2.698970 — 0.038511 
Log. p = 2.660459 

P = $457.57 = Cash payment. 

Note. — Such examples may be checked to prevent miscal- 
culations, by multiplying the result (the cash payment), by the 
tabular number given for corresponding number of years and 
percentage of interest in table on page 23 ; if calculations are 
correct, the product will be equal to the original bill. For in- 
stance, 457.57 X 1.092727 = 499.99909339 = $500.00. Thus, the 
calculation in the example is correct 

Sinking Funds and Savings. 

If a sum of money denoted by b, set apart or saved dur- 
ing each period of time, is put at compound interest at the end 
of each period, the amount will be : 

a = b at the end of the first period. 

a = b + br at the end of the second period. 

a = b -f br 4- br 2 at the end of the third period. 

At the end of n periods the last term in this geometrical 
series is br^~ x and the fi rst term is b, while the ratio is r. 
The sum of the series is the amount which according to the 
rules for geometrical progression (see page 69) will be : 

ribr*- 1 ) — b 

a = — - 

r— 1 

b (r n — 1) 
a = — i = 

r— 1 

Example. 

At the end of his first year's business a man sets apart 
$1200 for a sinking fund, which he invests at 4% per year. At 
the end of each succeeding year he sets apart $1200 which is 
invested at the same rate. What is the value of the sinking 
fund after 7 years of business ? 

Solution : 

1200 X (1.04 7 — 1) 

a = . . * ' 

1.04 — 1 

1200 X 0.31593 
a = 

0.04 
a = 39477.90 



LOGARITHMS. 85 



Example. 



A man 20 years old commences to save 25 cents every 
working day, and places this in a savings bank at 4% interest, 
computed semi-annually. How much will he have in the bank 
when he is 36 years old? (Note. — 25c. a day = $1.50 a 
week = 26 X $1.50 = $39 in six months. 4 % per year = 2 % 
per period of time ; 36 — 20 = 16 = 32 periods of time). 

Solution by formula: 

r — 1 

a = 39 X (1.02 32 — 1) 
1.02 — 1 

a == 39 X (1.8845 — 1) 
0.02 

a = 39 X 0.8845 X 50 

a = 1724.775 = #1724.77 = Amount. 

Thus, in 16 years a saving of 25c. a day amounts to $1724.77. 

If the money is paid in advance of the first period of time 
the terms will be : 

a = br at the end of the first period. 

a = br + br 2 at the end of the second period. 

a = br -j- br 2 -j- br B at the end of the third period. 

At the end of n years the last term in this geometrical 
series is br* and the first term is br, while the ratio is r. The 
sum of the series is the amount, which, according to rules for 
geometrical progressions (see page 69), will be : 

r (br n ) — br 

a = — : 

r — 1 
br (r n — 1) 



a 



r — 1 



Example. 



Assume that the man mentioned in previous example, 
instead of commencing to save money when 20 years old, 
already had $39 to put in the bank at 4 % the first period of 
time, and that he always kept up paying $39 in advance semi- 
annually. How much money would he then save in 16 years ? 



86 LOGARITHMS. 

Solution : 

a == 89 X 1.02 X (102 32 — 1) 
1.02 — 1 

39 X 1.02 X 0.8845 
a = — 



0.02 
a = 1759.27 

Thus, by paying the money in advance semi-annually, he 
will gain (1759.27 — 1724.77) = #34.50. 

If a principal denoted by ft is invested at a given rate of 
compound interest, and successive smaller or larger equal pay- 
ments denoted by b are made at the end of each period of time 
so that they will commence to draw interest at the beginning of 
the following period at the same rate as the principal, the 
formula will be : 

a = ft X r- + *fr n -l) 
r — 1 

but for logarithmic calculations it is more convenient to denote 
the rate of interest by y % and the formula will read : 

b (r n — 1) 
a=ft X r n + y 



ft X r n + 



100 

100 b (r n — 1) 



y 

ay — fty X r n 



r u _ 



100 


r n — 100 


ay 


+ 100 £ 


fty -f 100 b 


log. 


a y + 100 b 


P y -f 100 3 




log. r 


log. 


a y -f- 100 b 


P y -f- 100 b 



Log. t 

s 11 

Note. — Using these formulas it must be understood that 
n represents the number of periods of time that the ftrinciftal is 
invested, and that this first period is considered to be the period 
at the end of which the first payment, b, is made. 

Example. 

A man has $60 in a savings bank and he also puts in #25 
every month, which goes on interest every 6 months ; the bank 
pays 4% interest, computed semi-annually. How much money 



LOGARITHMS. 87 

can he save in 5 years in this way? (Note. — 4% per year = 
2% per 6 months, or per period of time, and $25 a month = $150 
every 6 months, or per period of time. The interest is computed 
Semi-annually ; therefore 5 years = 10 periods of time). 
Solution by formula : 

.' _ , 100£ (r n — 1) 

y 

* = 50 X 1.02- + 100 X 150 X (102^-1) 

a = 50 X 1.219 + 10 ° X 15 ° X °- 219 

2 

a = 60.95 + 1642.50 

a = #1703.45 = Amount. 

The original sum of $50 has increased to $60.95, and the 
monthly payments amounted to $1500. The last six payments 
did not draw any interest, as they were deposited in the last 
six months of the fifth year and would commence to draw inter- 
est at the beginning of the sixth year if the amount had not 
been withdrawn. 

Example. 

A man has $800 invested at 5%. How much must he save 
and invest at the same interest every year in order to increase it 
to $3000 in five years ? Interest is computed annually. 

Solution by formula : 

l, __ a y — P y X r u 

~ lOO r" — 100 

b _ 3000 X 5 — 800 X 5 X 1.05 5 

100 X 1.05 5 — 100 
h _ 15000 — 1.2763 X 4000 

100 X 1.2763 — 100 
, 15000 — 5105.2 



127.63 — 100 

9894.8 



27.63 

b = 358.118 = $358.12 to be paid in each year. 

The total payments will be : 

800 + 5 X 358.12 ±= $800 + $1790.60 = $2590.60. 

The rest of the amount is accumulated interest. The last pay- 
ment is made at the end of the fifth year ; therefore this money 
does not draw interest. 



88 logarithms. 

Example. 

A man calculates that if he had $1800 he would start 
in business. He has only $120, but is earning $15 a week and 
figures that he can save half of his weekly earnings. He puts 
his money in a savings bank, where it goes on interest every 
six months, at the rate of 4% a year. How many years will it 
take him to save the required amount? (Note. — $7.50 a week 
= 26 X 7^ = $195 in six months, and 4% per year = 2% per 
six months, or per period of time). 

Solution by formula : 

a y -f 100 b 



n = 



n = 



n = 



log. 



log. 



P j/ + 100 b 

log. r 

1800X2 + 100X195 
120 X 2 + 100 X 195 



log. 



log. 1.02 

3600 + 19500 
240 -4- 19500 



log. 1.02 

log. 1.1702 0.0683 Q . , e ^ , , x 

n = — £ = = 8 periods of time (nearly). 

log. 1.02 0.0086 \ y/ 

One period == 6 months ; 8 periods = 4 years ; therefore, 
under these conditions it takes four years to save this amount 
of money. 

If a certain sum of money is withdrawn instead of added, 
at the end of each period of time, the formula on page 86 will 
change to : 

, v n 100 b (r* — 1) 
a = p X r n — i ' 

y 

Every letter denotes the same value as it had in the formula 
on page 86, except that b represents the sum withdrawn instead 
of the sum added. 

Example. 

A man has $5000 invested at o% interest compounded an- 
nually, but at the end of each year he withdraws $200. How 
much money has he left after six years ? 
Solution : 

a = 1.05* X 5000 - 100 X 200 X (1.05* - 1) 

5 

a = 1.34 X 5000 - 10 ° X 200 X °' 34 

5 
a = 6700 — 1360 

a = $5340 = Amount. 



LOGARITHMS. 89 

If the deducted sum, b, exceeds the interest due at the first 
period of time, the amount a will become smaller than the prin- 
cipal^, and in time the whole principal will be used up. This 
will be when : 

100 £ (r n — 1) 
P X r" = 

This transposes to 

100 £ 

rD= 100 6— py 

100 b 



log. 



100 b — py 

n = Jo~g~? 

Example. 

A principal of $5000 is invested at 4% per year, but at the 
end of each year #600 is withdrawn. How long will it take to 
use the whole principal? 

100 x 600 
™8- 100 X 600 — 5000 X 4 



n = 



log. 



71 



n 



log. 1.04 

60000 
60000 — 20000 



~~ log. 1.04 

_ log. 1.50 
log. 1.04 
^ = 0176091 
0.017033 
n = 10.3 years. 

Paying a Debt by Instalments. 

This same formula applies also in this case; for instance: 
A man uses $1500 every year toward paying a debt of $10,000, 
and h% interest per year. How long will it take to pay it? 

100 x 1500 
l°g- 100 X 1500 — 10000 X 5 



n = 



n = 



n = 



n = 



log. 



log. 1.05 

150000 
100000 



log. 1.05 
log. 1.5 
log. 1.05 
0.176091 



0.021189 
n = 8.3 years. 



9° 



LOGARITHMS. 



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LOGARITHMS. 1 09 



lO lO >0 lO lO JO JO HH rt< Tfl -* H^ H/l t}H Ttf CO CO CO CO CO CO CO CO CM <M 
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tH CO "^ CO i— I GO H^ GO CM CO ""* CO CM CO HH OOtHCO^CO (N O Tt< CI (M 

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H 00 CO 00 H GO JO CO O CO ifl CO O MO C^Oit-Tf-i Oi CD CO O GO 

HHINCOTf t^ O CO 1— r— GO O O O iH CM CM CO Hi o JO CD t— CO GO 

co co co co co cDcocococo co co r— r— r- r- r— r— r- r- r— r— r— r— r- 

r— r— t— r— r— r— r— r— r— r— r— r— £—i— ir- r— r— r— r— r— r— r— r— r— r— 



00 



»o co o 00 r- -^o-jHr~C3 00x100 oodoho o cm co o 

cm r- cm r— cm r~(Mcoo-ti oimoo^ r— o hi r- o (noccoco 

OMONO r— JO CM O r— H^ CM Oi r— H^ h010«H GO JO CM O r— 

th-i— (<Mco— ^ — jHocor~r~ co jiqoh cm cm co h^ »o jo co r— go go 

cococococo cococococo cococor~r~ r~r~r~r~r~ r-r-r-r-r- 

r~r~r~r~r~ r~r~r~r~r~ r~r~r~r~r~ r~r~r~r~r~ r~r~r~r~r~ 



CNMWN O lO O M 1(5 CO CD H^ _i r- CM JO r— CO GO CO CO Oi H^ GO 

JO O JO O JO Oi^QMr- 1— l J.O Oi CO O OCOCOOKM JO GO O CO JO 

cir~-tH<MCs co h/i th o co Tti-Hoococo -hgoocmo r— hi cm co 

oh'nmm ~^xococor~ oodooh (nn wfic iocor~r~cc 

co co co co co cococococo co co co r— r- r~r~r~r~t~ r~r~r~t~r~ 

r~r~r~r~t~ r~r~r~r~r~ r~r~r~r~r~ r~r~r~r~r~ r~r~r~r~r~ 



50 



jo r- go go r— -t< o >ooh cm cm o r- co go cm hi 10 10 co ih r- cm jo 

r- cm r- cm r- cm r— ih jo o h^ go cm jo Oi (NOojino go ih co co go 

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o -i— 1 cm co co -* io co co r- oooiOJ Oh cm cm co -* h^ jo co r- r- co 

cococococo cococococo co co co t— r— r~ r— r— r— r— r— r— r— r— r— 

t~r~r~r~r~ r~r~r~r~r~ t~r~r~r~r~ r~r~r~r~r~ r~r~r~r~r~ 



10 



Oi CM CO CO CM O0-H»0r~ OOGOCO-^O »O00H(MiM O GO HI Oi CO 

Oi JO O JO O jO Oi HI GO CM CD O H^ GO CM JO CO CM JO GO -HCOCCCOrH 

r- jo co o co 10 « o i- o cm o r- h^ cm Cico-^-nco co co or- 10 

OHjqcoco -jiiococor- 00 c o o h t—icnco-^hh iocor-r-00 

cococococo cOcocdcdcd co cO co r— r— r— r— c— r— r— r— r— r— r— r— 

r— r— r- r— r— r- r— t— r- r- r— r- c- r- r— r— r- r— r— r— r— r-r-r-r— 



hi r— go go r- jo th co o co ^-*coocd hiooooj© ooioni-h 

<n t- (M r- 01 r-oqcoHio os co r- th h^ go — < -r r- o co co a h ^ 

r-^<Mor- -^ cm oi r— -* hoco-^t-i gocdcoogo jo cm os r- h^ 

OH(N(MfO H^ jC3 JO CO Jt- GO GO Oi O tH HtNCO^Tti JO CO CO JC- GO 

cococococo cococococo cococo r— r— r— r— r— r— r— r— r— r— r— r— 

r-r-r-r-r— r-r-r-r-r- r-r-r-r-r- r-r-r— r-r— r-r-r-r-r- 



M 



OS CM CO ^t< CO OMNCDO O O Oi CO CO GO CM H^ CO CO JO CO OS H/t GO 

ioooio o^oicor- (N co o co r- o ^ i- o co co Oi th h^ co 

CO-^t— (OiCO -^ h CO CC CO HOOiOCOO 00 JO CM O r— TfrHOCCCO 

oh cqtNoo H^oiocor- goqoojoh himco^^ jo co o r- go 

cococococo cococococo co co co r— r— r— r— r— r— r— r— r— r— r— r— 

r-r-r— r-r— r-r-r-r-r- r— r-r-r-r- r-r-r— r-r— r-r-r-r-r- 





CO CO CO CS GO 


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GO 


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pp 


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27 ""— ' *"* — *-7 23 


27 r^ 


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71 71 71 71 


71 71 71 71 71 


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Logarithms. 113 



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■^ji ^hS **jti ^hH "^^ 


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co co co 


co co 


CO CO CO CO CM 


(M CM CM (M Ol 


(OOOCDO 


CO O O CC O 


CO CC O O O 


co co co co co 


co co 




»o O O iS CO 


T— 1 t— CO t— — 1 


-I-lOC 


co ta 


COHt-Nt- 


O CO O CO CO 




NCOi'CO 


03 lO Ci (M CO 


Ci CM O 


cc —1 


-fi l- Ci CM -* 


r- Ci t— 1 co 10 


© 


lOHOOTfO 


t— CO Ci CO CM 


CCiO- 


1 — f 


O CO CM Ci l-O 


H 1- ■* O © 


O *-■ 1-1 CM CO 


CO -f -t- 10 CO 


co r- x 


X Ci 


OOhhC( 


COCOtJUOiO 




c© c© co co co 


CO CO CO CO CO 


CO CO CO 


CO CO 


-t -t -t -+ -+l 


tHH -^HH -^H -^1 -^Jl 




00 GO CC 00 00 


cc cc cc cc cc 


66 66 66 


X X 


cc cc cc cc cc 


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O 0<J CM — i O 


i f- Ci -+ t- 


cm CO 


CO CM 


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X T-H CO ^tf ^ 




00-*CCiN 


10 r. m c c 


CO CC Ci 


0<l »o 


x co cc x 


OcOiOt-O 


00 


TtHt-no 


O CM Ci O — < 


X -* 


t- CO 


Ci CO CM X ^ 


i-H t- CO Ci- 10 


OHr-ljq CO 


co -* Tt< »c co 


CO t- X 


X Ci- 


O O H rH (N 


CO CO Tf tH 10 




co co co co co 


CO CO CO CO CO 


CO CO CO 


CO CO 


CO ^ -t -v -+i 


"^ -+ "^ "^ "^t* 




00 00 00 00 00 


cc cc cc cc cc 


XXX 


X X 


cc ZC cc cc cc 


cc cc cc cc cc 




r- 00 t- 


moo©-* 


t-Ci O 


CO Ci 


cc CM ir- cm 


O Ci H <M <M 




CO CO Ir- t-i 


Ci- co co 


CC Ci CO 


cc x 


1-1 -^ r- Ci cm 


^HH CO Ci T-H CO 




CO O CC CO Ci 


cm x »ra -h 


t- CO 


CC CM 


Ci l-O 1-H t- Tt 


OO CM-Ci O 


r* 


O -H ^H <M (M 


CO ^f ^ "CO CO 


CO 1— X 


X Ci 


Ci O — 1-1 CM 


CO CO -t -* 




CO CO CO CO CO 


CO CO CO CO 00 


CC CO CO 


CO CO 


cc -t -t -t -* 


"*HH 'HH -^H T^l *^1 




GO X X 00 GO 


X X X X X 


CO X X 


X X 


cc cc cc cc cc 


CC CC CC cc cc 




<M CO -rJH CO CM 


OO CNIt-H 


r^ CO f- 


r— ir- 


JO 03 Ci O O 


CO CO 




CO t— 1— 1 u© Ci 


00 CO O CO t- 


O CO cc 


Ci CM 


10 x co co 


X O CM lO £- 





CO Oi C O) X 


O —1 X ^f 


r- co cc. 


IQ CM 


X Tj< T-I t- CO 


Ci CO CM X rt< 


OOhNN 


co -<* -* »o co 


cc t- 1— 


X Ci 


Ci O ^h — CM 


CM W*-*iO 




co co co co co 


CO CO CO CO CO 


CO CO cc 


CO CO 


CO -t -t "* ^* 


"^ tHH *tHH "^ tHH 




00 00 X CC X 


cc cc cc cc cc 


XXX 


X X 


cc cc cc cc cc 


cc cc cc cc cc 




go c co 


CO CO Ci CO t— 


T-H CO -f "^ Ttl 


CM O CO CM t- 


T-H T^ O X X 




O O X CM 


CC O CO I- 


-? I- 


CO cc 


Ci cm ■* r- ci 


CM ^ CO X O 


US 


CM Ci lO — i X 


-* - 1- CO c 


CO "M Ci 


tO i—l 


1 1^ O CC CM 


Ci t-h ir- -* 


OOhc^J) 


co -+-foc 


CO t— 1- 


X Ci 


© O rt rt N 


CM CO T^ Ttl lO 




CO CO CO CO CO 


CO CO CO CO CO 


CO CO CO 


CO CO 


CO Tt< -JH -* -fl 


^T ^^ ^J^ *^^ ^ST 




CC CC CC CC CC 


cc cc cc -cc x 


cc Ic cc cc cc 


X.X XXX 


CC CC CC CC cc 




-^ c© co ^ 


CM Ci i©> O -+i 


t~ — 1 


T— 1 T— 1 


Ci t fi O >0 


Ci CM -* CO CO 




O -* QO (N CD 


O CO 1- 1 f 


t- 1-1 "* 


t- CO 


Oi k-O X t- CO 


lO X O (M -¥ 




CN CO ^ rH 1- 


-t cc co Ci 


O CM X 


-tf — 1 


£- CO Ci CC CM 


cc •*& H t— CO 


*# 


OOh^-N 


co -t ■* c 


cc r- r— 


X Ci 


Ci CO O t-h CM 


CM CO tjh ^ »0 




CO CO CO CO CO 


CO CO CO CO CO 


~~. ^ ^ 


CO CO 


CO -f -* -HH -t 


"^7" ^^ ^P ^^ ^^ 




cc cc cc cc cc 


06 x 66 x co 


XXX 


X X 


cc cc cc ~c cc 


cc cc cc cc cc 




Ci i—l CM CM O 


00 10 t— i r- 1— 1 


-* CO X 


X X 


t f -h t- CM 


OOOliiTji 




CO X CM CC O 


CO L— — -*t X 


1-1 -r i— 


O CO 


CO Ci CM -f L— 


Ci- CM ^ CO X 


CO 


tH t— ^* O 1— 


co Ci co 01 x 


»o — 1 t- *># 


CO CM Ci lO 1-1 


t-^OCOCN 


O O i— CM CM 


CO CO — t* >o >o 


CC L~— 1— 


X Ci 


Ci O O t-h CM 


CM CO "* "<* iC 




co co co co co 


CO CO CO CO CO 


CO CO CO 


qq f^fy 


CO -t -f -& ^ 


"^i '^^H "^1 ^^ ""^^ 




00 00 XX) 00 CO 


X X X X X 


CC CC ZC cc cc 


X X X X X 


cc cc cc cc cc 




1- X X t- 


lO CM X CO t— 


,_, ^-> ,^v 


>o 


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•*$ I— O t-h CM 




b rt O C CO 


t- .-1 t# CO tH 


O X 1"H 


-t 1- 


O CO iO X — 1 


CO »o X O CM 


N 


O t- CO c c 


CM C I.O rH X 


-t 1- 


CO Ci 


COl X •* H 


r- co Ci co cm 


— — ' 01 


CO CO -t "O lO 


CC 1- t- 


X X 


Ci O CO -T-i CM 


CM CO CO ^ O 




CO CO CO CO CO 


CO CO CO CO CO 


CO CO CO 


cc CO 


CO ~t -+ -* "t 


~"f -H^ -Hri "rt* T^l 




cc cc cc cc cc 


cc cc co cc x> 


X X X 


X X 


cc cc 'cc co cc 


cc cc cc cc cc 




t-i co -f -t co 


T* CC ~f © -+ 


NOCM 


O? CM 


— 1 Ci CO CM i— 


(MiOXffiO 




h 10 c co i- 


i— 1 -+ X CM O 


X CM lO 


X 1-1 


-t CO Ci CM ^ 


r- Ci T-t co co 


H 


O CO CM Ci lO 


(M X -f t-i t- 


CO O CO 


CM Ci 


lO H 1- •* O 


CO CM Ci iO tH 


OOr-HOl 


CO CO "Tf »o iO 


CC L— 1— 


X X 


Ci O O 1— 1 CM 


CM CO CO -^ »c 




co ^^ C^ CO CO 


CO CO CO CC CO 


CO CO CO 


-v- OC 


cc -r -f -* ~+ 


^TJ^ ^Jjl '^rj^ "^ijl ^r 




X 00 66 X 66 


X X X X X 


X X X 


X X 


cc cc cc cc cc 


cc cc cc cc cc 




t- Oi O O Ci 


r fHC h 


-t t- X 


Ci Ci 


CC CC CO Ci 


co i©> r- x 




■* 00 CO 1- 


-f X CM iO Ci 


CM uO X 


— J- 


1- co co x 


co r— Ci 


O 


a >o <m x 


t-i r f co cc 


CO Ci O 


CM X 


Tf 1— 1 i— CO Ci 


CO CM X Tfl O 


Ci O T— 1 T— 1 <M 


co co^t 10 >o 


CC CC t- X X 


Ci CO O h 1— 1 


CM CO CO Tf lO 




CM CO CO CO CO 


CO CO C* CC CO 


CO CO CO 


CO CO 


CO -v "* ^t< ^ 


Tfi ^ ~rH t^ "^jn 




CC CC CC cc cc 


66 x x x x 


XXX 


X X 


cc cc cc cc cc 


cc cc cc cc cc 




O I* JC © © 


1-1 N CO <* »© 


co r» x 


Ci c 


rH N CO rj* L<5 


CO ^ X Ci © 


fc 


r~ r- i- i- X 


X X X X X 


XXX 


x en 


00000 


00000 


C© CO CO C© C© 


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CO CO CO 


CO CO 


CO CO CO CO CO 


CO CO CO CO I- 



ti4 



LOGARITHMS. 



V | OI OI <M OI O} 

H 1 © © © © © 


1— I T— 1 1— t —H H 


1— I 1— 1 1— 1 1— 1 1— 1 


TH T-i' © © © 


© © © © © 


CO CO CD CD CO 


CO CO CD CD CO 


© © © © © 


© © © © © 




O-^rH XCO 


CONiOl^O 


Ci Ci 1— WO OI 


© -f x oq wo 


t- X © X t- 




t- © ■>— i oj if 


WO t- X Oi O 


t-i 04 CO if WO 


wo © © t- t- 


1— 1- 1— t- t- 


© 


(MQOlOHt- 


CO Oi WO r- 1 X 


-f O CO OI X 


if © © OI X 


if © © oq x 


CD CO t- GO GO 


Oi Oi O t— i t— ( 


oq co co if -f 


wo © © t- t- 


X © © © © 




if if if if -f 


• if if iO W0 WO 


WO WO WO WO iO 


W0 WO WO >o »o 


iO WO wo © © 




GO GO GO GO GO 


X X X X X 


X X X X X 


X X X X X 


X X X X X 




00 (M OS O <M 


t >-*cr- 


COQOMQN 


X CO X OI WO 


t- X © X t- 




t-I CO if © X 


Ci — ■ Oq CO if 


WO CD t- X Oi 


© © © T-i T-i 


1— 1 -H iH i— 1 ^H 


00 


!N 00 ^ O C 


oi © wo i— 1 1— 


co Ci io — > t- 


co © © oi x 


if © © cq x 

X © © © © 


CO CO t- GO GO 


Oi Oi O T-H t— I 


oi oi co if -f 


WO CO © t- t- 




"^ "^ if "^ if 


if if WO W0 WO 


W0 WO WO W0 W0 


O WO WO WO wo 


WO WO WO © © 




GO GO GO GO GO 


X X X X X 


X X X X X 


X X X X X 


X X X X X 




i-l O GO -^ o 


WO Oi OI WO CO 


t-t- CO if T-I 


t- CO X OI WO 


t- X © X t- 




WO t- X O OI 


co -rji co t- x 


Oi O t-i OI CO 


CO if if wo wo 


WO WO WO ifO wo 




t— i t— CO © © 


!N X ^ O O 


CM Ci W0 — i t- 


CO © WO T-i t— 


co © »o — t- 


2^ 


© © t- X X 


O O O H H 


oi oi co -f if 


O WO © 1- t- 


X X © © © 




"■^H "^^ "^rp "^r ""T^ 


~f -f wo wo wo 


W0 WO WO WO WO 


O WO WO WO wo 


wo wo >o © © 




GO GO GO GO GO 


X X X X X 


X X X X X 


X X X X X 


X X X X X 




O X © CO © 


■*XH^>C 


CO CO W0 CO © 


t— oq t- t-i if 


t- X © X t— 




^OM ^flO 


i- X O T-i CM 


CO ~f WO © t- 


t- X X © © 


Ci © © © © 


55 


O t- CO Oi o 


t-h t- if o CO 


Ol X if © © 


OI X if © © 


OI X if © © 


© © r- i— x 


Oi Oi O t— ( i— ( 


OI OI CO if if 


WO W0 © 1— 1— 


X X © © © 




if if rjn if if 


if -t O lC O 


WO WO WO WO W0 


WO WO WO WO wo 


wo wo >c © © 





GO GO X X GO 


X X X X X 


X X X X X 


X X X X X 


X X X X X 


X CO -f h t- 


OI CD O OI if 


WO W0 if C<l © 


© oq t 1 if 


© X © X t- 




oi -f co x © 


H OI f WO CO 


I— X © © T-I 


i-i OI OI CO CO 


co co co co co 


10 


© © oi x -f 


t— t- co Ci wo 


T— 1 t— CO © © 


OI X if © © 


OI X if © © 


co co t- t- x 


Oi Oi O O t-i 


oi oq co -f if 


W0 iO © 1- t- 


X X © © © 




if if if TT if 


if if WO W0 WO 


WO WO WO WO W0 


wo wo wo wo wo 


wo wo wo © © 




X X X X X 


X X X X X 


X X X X X 


X X X X X 


X X X X X 






CD >0 (N O O 


t— l WO Oi h CO 


if Tf CO T-H © 


© 1-1 © © if 


© X © X X 




CD X O —i CO 


WO CD t- Oi O 


t-i oi co if if 


wo © © t- t- 


t- t- t- t- t- 


<# 


Oi »o co x if 


O CD OI X WO 


l— 1 1— CO © wo 


i— 1 1- co © wo 


T-t t— CO © wo 


W0 CO t- t- X 


Oi Oi O O t— i 


oq oq co co if 


W0 WO © © t- 


X X © © © 




■^ if if "if if 


if if WO WO wo 


WO WO wo wo wo 


wo o wo wo wo 


wo wo wo wo © 




X X X X X 


X X X X X 


X X X X X 


X X X X X 


X X X X X 




if CO — ' x if 


Oi if I— O OI 


co co oi i—i x 


WO i— 1 © © CO 


© X X X X 




o oj if wo t- 


X O T-t CO if 


WO © t- X X 


Ci © © H T— 1 


T— 1 T— 1 T— 1 T— 1 T— 1 




Oi wo i— i t— co 


ci co oi x if 


© co oi x if 


© t— CO © WO 


t— I t— CO © W0 


CO 


WO CO t- t- X 


X Oi O O — i 


CM OI CO CO ~f 


wo o © © t- 


X X © © © 




■*f if if if if 


if if WO W0 WO 


wo wo wo wo wo 


wo wo wo o wo 


WO WO O WO © 


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X X X X X 


X X X X X 


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X X X X X 


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Oq ~H Ci CO oi 


X OI CO Oi t-i 


OI OI t-i © X 


if © WO © CO 


© t- X © X 




if CO t- Oi t-i 


OI if WO CD X 


© © ii OI OI 


CO if -f W0 WO 


wo wo wo wo wo 


N 


x if o co co 


Oi wo — I- CO 


© © OI X -f 


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wo © t- t— x 


X Ci O O i-i 


t— I "M CO CO if 


WO WO © © t— 


X X © © © 




■*~^ ^^ TT *^r ^^ 


if if WO WO wo 


wo wo »o wo wo 


WO WO WO WO WO 


W0 W0 WO WO © 




X X X X X 


X X X X X 


X X X X X 


X X X X X 


X X X X X 




OSl--*H 


OHCt-O 


t-h t— i © Ci t— 


if © WO © CO 


WO t- X © X 




X Oi T-i co wo 


CO X Oi O O} 


CO if WO W0 © 


t— X X X © 


Ci © © © © 




1- CO O CO OI 


X -f O 1- CO 


Ci WO t-i t- CO 


© W0 i-i X- CO 


© WO t-i t- CO 


iH 


iO "O t- 1- X 


X Oi O CO' H 


t— I OI CO CO -f 


if wO © © t- 


1- X © © © 




-f if if if if 


if if wo wo wo 


lO wo wo wo wo 


W0 WO WO W0 WO 


WO WO WO WO © 




X X X X X 


X X X X X 


X X X X X 


X X X X X 


X X X X X 


CO t— WO CO © 


wo Oi CO CO X 


© © © X © 


co © if © oi 


WO t- X © X 


i-i CO WO 1- X 


O t— i CO if o 


t- X © © © 


T-I T-I OI OI CO 


CO CO CO CO CO 


„ ! 1- CO © WO t-i 


x if o cd oa 


X if © © CO 


Ci W0 i— I t— CO 


© W0 -i t- CO 


© i.O S O 1- CO 


X Oi O O t-i 


t-i oi co co if 


if wo © © t— 


t- X © © © 


if -f if -f -f 


if if wo o wo 


wo wo wo wo wo 


wo wo wo wo wo 


WO WO WO WO CO 


GO GO X GO GO 


X X X X X 


X X X X X 


X X X X X 


X X X X X 




H N M ■* M5 


(ObOCftO 


rt « M t# 15 


© r- 00 © © 


H « eo ^ M5 


£ 


o © c c © 


© © © © H 


H tH H H i-l 


iH H H i-l « 


« N N « (M 


t- s- r- *- r» 


I- *i *i I- i- 


i- *• r- r- *- 


i' ii ii li Ji 


r- *- Ji r- r- 



LOGARITHMS. 



"5 



Q 


O O OS 


OS OS OS OS OS 


OS OS OS OS OS 


es 00 co 00 00 


00 CO CO CO CO 


coo c »o 


iO O O iO iO 


1O1OO1O1O 


iO »0 lO 


\o »o 10 *o >o 




lO MOO WOO 


(M O £- CO OS 


OS CO O CO 


lO O »0 00 H 


CM ^ Tt* CO CM 




r— ir- »o 


lO^MtNH 


O OS GO t- O 


-»* CO t-i OS 00 


CO-*(MOGO 


© 


i<OCD(N(X) 


^OtCIMQO 


-tf CS iO 1— 1 tr- 


co os 10 


CM 00 "* O »0 


HNMCOCO 


-** »o >o 


ie- r- co os os 


O O T-H CM CM 


CO CO Tf »C id 




O^OOO 


OOOOO 


OOOOO 


Ir- I— Jr- t- Jr- 


Jr- ir- Jr- t- Ir- 




OC GO 00 00 00 


CO CO CO CO CO 


CO 00 GO CO 00 


00 00 GO GO GO 


00 GO CO CO CO 




kO CM 00 ^ O 


cm >o co os 


© OS Jr- -* — 1 


Jr- cm O (M 


Tjl lfi CD lO Tjl 




1— 1 t-h O O OS 


OS GO t- O O 


O CO <M T-H O 


00 Jr- -* cm 


O CO O -^ CM 


00 


"* O O CM t- 


tJffllflHt" 


co os 10 t— 1 1— 


CM CO "tf O O 


CM Jr- CO CS iQ 


H^<NCOM 


T*l -r* lO O O 


i— IT- GO CS OS 


O O tH CM CM 


CO CO ^ ^ 




OOOOO 


OOOOO 


OOOOO 


Jr- t- Jr- Jr- Jr- 


Ir- Jr- ir- Jr- t- 




CO GO CO CO 00 


CO 00 00 GO GO 


GO CO CO CO GO 


00 GO GO CO CO 


00 CO GO Cfe CO 




OMO-^ffi 


co os t-i 


tH O 00 O CM 


CO CO CO t-i ^ 


O ir- CO 00 O 




10 »o •<* ■>* CO 


«(NhhO 


OS CO O O ^* 


CM tH CS 00 O 


■*# CM O GO O 


jr- 


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co os to t-i t— 


CM CO "^ O O 


CM 00 CO CS O 


t-h ir- co 00 ~r 


HHMCOC5 


T*< T* lO O O 


Jr- Jr- GO OS CS 


O O tH t-i CM 


CO CO Tf Tt 




OOOOO 


OOOOO 


OOOOO 


Jr- Jr- Jr- Jr- Jr- 


Ir- Ir- Ir- Jr- Jr- 




00 GO CO CO 00 


CO CO CO CO CO 


CO 00 GO CO GO 


00 GO 00 CO CO 


00 GO CO CO CO 




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126 



HYPERBOLIC LOGARITHMS. 



HYPERBOLIC LOGARITHMS. 

The hyperbolic or Napierian logarithm of any number may 
be obtained by multiplying the common logarithm by the con- 
stant 2.302585; practically 2.3. 

Table No. 6 gives the hyperbolic logarithms from 1.01 to 
30. The hyperbolic logarithm of numbers intermediate between 
those which are given in the table may be obtained by inter- 
polating proportional differences. 



TABLE No. 6. — Hyperbolic or Napierian Logarithms 
of Numbers. 



N 


Log. 


N 


Log. 


N 


Log. 


N 


Log. 


1.01 


0099 


1.26 


2311 


1.51 


0.4121 


1.76 


0.5653 


l.(>2 


0.0198 


1.27 


0.2390 


1.52 


0.4187 


1.77 


0.5710 


1.03 


0.0 96 


1.28 


0.2469 


153 


0.4253 


1.78 


0.5766 


1.04 


0:592 


1.29 


0.2546 


1.54 


0.4318 


1.79 


0.5822 


1.05 


0488 


1.30 


0.2624 


1.55 


0.4383 


1.80 


0.5878 


1.06 


0.0583 


1.31 


2700 


1.56 


0.4447 


1.81 


0.5933 


1.07 


0.0677 


1.32 


0.2776 


1.57 


0.4511 


1.82 


05988 


1.08 


00770 


1.33 


0.2S52 


1.58 


0.4574 


1.83 


0.6043 


1.09 


0862 


1.34 


0.2927 


1.59 


0.4637 


1.84 


0.6098 


1.10 


0.0953 


1.35 


3001 


1.60 


0.4700 


1.85 


0.6152 


1.11 


0.1044 


1.36 


0.3075 


1.61 


0.4762 


1.86 


0.6206 


1.12 


0.1133 


1.37 


0.3148 


1.62 


0.4824 


1.87 


0.6259 


1.13 


0.1222 


1.38 


0.3221 


1.63 


0.4886 


1.88 


0.6313 


1.14 


01310 


1.39 


3203 


1.64 


0.4947 


1.89 


0.6366 


1.15 


0.1398 


1.40 


0.3365 


1.65 


0.5008 


1.90 


0.6419 


1.16 


0.1484 


1.41 


0.3436 


1.66 


0.5068 


1.91 


0.6471 


1.17 


0.1570 


1.42 


0.3507 


1.67 


0.5128 


1.92 


06523 


1.18 


0.1655 


1.43 


0.3577 


1.68 


0.5188 


1.93 


0.6575 


1.19 


0.1740 


1.44 


0.3646 


1.69 


05247 


1.94 


0.6627 


1.20 


0.1823 


1.45 


0.3716 


1.70 


0.5306 


1.95 


0.6678 


1.21 


0.1906 


1.46 


0-3784 


1.71 


0.5365 


1.96 


0.6729 


1.22 


0.1988 


1.47 


0.3853 


1.72 


0.5423 


1.97 


0.6780 


1.23 


0,2070 


1.48 


0.3920 


1.73 


0.5481 


1.98 


0.6831 


1.24 


0.2151 


1,49 


3988 


1.74 


0.5539 


199 


0.6881 


1.25 

bssjrjcese: 


0.2231 


1.50 


0.4055 


1.75 


0.5596 


2.00 


0.6931 



HYPERBOLIC LOGARITHMS. 



127 



■ N 


Log. 


N 


Log. 


N 


Log. 


N 


Log. 


•2.01 


0.6981 


2.41 


0.8706 


2.81 


1.0332 


3.21 


1.1663 


2.02 


0.7031 


2.42 


0.883S 


2 82 


1.0367 


3.22 


1.16'.<4 


2 03 


0.70S0 


2.43 


0>S79 


2.83 


10403 


3.23 


1.1725 


2.04 


0.7129 


244 


0.8920 


2.84 


1.0438 


3.24 


1.1756 


2.05 


0.7178 


2.45 


0.8961 


2.85 


1.0473 


3.25 


1.1787 


2.06 


0.7227 


2.46 


0.9002 


2.86 


1.0508 


3.26 


1.1817 


2.07 


0.7275 


2.47 


0.9042 


2.87 


1.0543 


3.27 


1.1848 


2.08 


0.7324 


2.48 


0.9083 


2.88 


1.0578 


3.28 


1.1878 


2.09 


0.7372 


2.49 


0.9123 


2.89 


1.0613 


3.29 


1.1909 


2.10 


0.7419 


2.50 


0.9163 


2.90 


1.0647 


3.30 


1.1939 


2.11 


0.7467 


2.51 


0.9203 


2.91 


1.0682 


3.31 


1.1969 


2.12 


0.7514 


2.52 


0.9243 


2.92 


1.0716 


3 32 


1.2000 


2.13 


0.7561 


2.63 


0.9282 


2.93 


10750 


3.33 


1.2030 


2.14 


0.7608 


2 54 


0.9322 


2.94 


1.0784 


3.34 


1.2060 


2.15 


0.7655 


2.55 


9361 


2.95 


1.0818 


3.35 


1.2090 


2.16 


0.7701 


2.56 


0.9400 


2.96 


1.0852 


3.36 


1.2119 


2.17 


0.7747 


2.57 


0.9439 


2.97 


1.0886 


3.37 


1.2149 


2.18 


7793 


2.58 


0.9478 


2.98 


1.0919 


3.38 


1.2179 


2.19 


0.7839 


2.59 


0.9517 


2.99 


1.0953 


3.39 


1.2208 


2.20 


0.7885 


260 


0.9555 


3. 


1.0986 


3.40 


1.2238 


2.21 


0.7930 


261 


0.9594 


3.01 


1.1019 


3.41 


1.2267 


2.22 


0.7975 


262 


0.9632 


3.02 


1.1053 


3.42 


1.2296 


2.23 


0.8020 


263 


0.9670 


3.03 


1.1086 


3.43 


1.2326 


2.24 


0.8065 


264 


0.9708 


3.04 


1 1119 


3.44 


1.2355 


2.25 


0.8109 


265 


0.9746 


3.05 


1.1151 


3.45 


1.2384 


2.26 


0.8154 


2.66 


0.9783 


3.06 


1.1184 


3.46 


1.2413 


2.27 


0.8198 


2.67 


9821 


3.07 


1.1217 


3.47 


1.2442 


2.28 


0.8242 


2.68 


0.9858 


3.08 


1.1249 


3.48 


1.2)70 


2.29 


0.8286 


2.69 


0.9895 


3.09 


i.r_82 


3.49 


1.2499 


2.30 


0.8329 


2.70 


0.9933 


3.10 


1.1314 


3.50 


1.2528 


2.31 


0.8372 


2.71 


0.9969 


3.11 


1.1346 


3.51 


1.2556 


2.32 


0.8416 


2.72 


1.0006 


3.12 


1.1378 


3.62 


1.2585 


2.33 


0.8459 


2.73 


1.0043 


3.13 


1.1410 


3.53 


1.2613 


2.34 


0.8502 


2.74 


1.0080 


3.14 


1.1442 


3.54 


1.2641 ' 


2.35 


0.8544 


2.75 


1.0116 


3.15 


1.1474 


3.55 


1.2669 


2.36 


0.8587 


2.76 


1.0152 


3.16 


1.1506 


3.56 


1.2698 


2.37 


0.8629 


2.77 


1.0188 


3.17 


1.1537 


3.57 


1.2726 


2.38 


0.8671 


2.78 


1.0225 


3.18 


1.1569 


3.58 


1.2754 


2.39 


0.8713 


2.79 


1.0260 


3.19 


1.1600 


3.59 


1.2782 


2.40 


0.8755 


2.80 


1.0296 1 


3.20 


1.1632 


3.60 


1.2809 



128 



HYPERBOLIC LOGARITHMS. 



N 


Log. 


N 


Log. 


N 


Log. 


N 


Log. 


3.61 


1.2837 


4.01 


1.3888 


4.41 


1.4839 


481 


1.5707 


3.62 


1.2865 1 


4.02 


1.3913 


4.42 


1.4861 


4.82 


1.5728 


3.63 


1 2892 : 


4.03 


1.3938 


4.43 


1.4884 ! 


4.83 


1.5748 


3 64 


1.2920 


4.04 


1.3962 


4.44 


1.4907 


4.84 


1.5769 


3.65 


1.2947 


4.05 


1.3987 


4.45 


1.4929 


4.85 


1.5790 


3.66 


1.2975 ; 


4.06 


1.4012 


4.46 


1.4951 


4.86 


1.5810 


3.67 


1.3002 


4.07 


1.4036 


4.47 


1.4974 


4.87 


1.5831 


3.68 


1.3029 ! 


4.08 


1.4061 


4.48 


1.4996 


4 88 


1.5851 


3.69 


1.3056 j 


4.09 


1.4085 


4.49 


1.5019 


489 


1 5372 


3.70 


1.3083 


4.10 


1.4110 


4 50 


1.5041 


4.90 


1.5892 


3.71 


1.3110 


4.11 


1.4134 


4.51 


1.5063 


4.91 


1.5913 


3.72 


1.3137 


4.12 


1.4159 


4.52 


1.5085 


4 92 


1.5933 


3.73 


1.3164 


4.13 


1.4183 


4.53 


1.5107 


4 93 


1.5953 


3.74 


1.3191 ! 


414 


1.4207 


4.54 


1.5129 


4 94 


1.5974 


3.75 


1.3218 


4.15 


1.4231 


4.55 


1.5151 ! 


4.95 


1.5994 


3.76 


1.3244 


4.16 


1.4255 


4.56 


1.5173 


4 96 


1.6014 


3.77 


1.3271 


4.17 


1.4279 


4.57 


1.5195 


4.97 


1 6034 


3.78 


1.3297 


4.18 


1.4303 


4.58 


1.5217 


i 4.98 


1.6054 


3.79 


1.3324 ! 


4.19 


1.4327 


4.59 


1.5239 


4.99 


1.6074 


3.80 


1.3350 


4.20 


1.4351 


4.60 


1.5261 


5. 


1.6094 


3.81 


1.3376 


4.21 


1.4375 


4.61 


1.5282 


5.01 


1.6114 


3 82 


1.3403 ; 


4.22 


1.4398 


4.62 


1.5304 


5.02 


1.6134 


3.83 


1.3429 


4.23 


1.4422 


4.63 


1.5326 


5.03 


1.6154 


3.84 


1.3455 


4.24 


1.4446 


4.64 


1.5347 


' 5.04 


1.6174 


3.85 


1.3481 


4.25 


1.4469 


4.65 


1.5369 


5.05 


1.6194 


3.86 


1.3507 


4.26 


1.4493 


4.66 


1.5390 


5.06 


1.6214 


3.87 


1.3533 


4.27 


1.4516 


4.67 


1.5412 


5.07 


1.6233 


3 88 


1.3558 


4.28 


1.4540 


4.68 


1.54 33 


5.08 


1.6253 


3.89 


1.3584 


4.29 


1.4563 


4.69 


1.5454 


5.09 


1.6273 


3.90 


1.3610 


4.30 


1.4586 


4.70 


1.5476 


5.10 


1.6292 


3.91 


1.3635 


4 31 


1.4609 


4.71 


1.5497 


5.11 


1.6312 


3.92 


1.3661 


4.32 


1 .4633 


4.72 


1 5518 


5.12 


1.6332 


3.93 


1.3686 


4.33 


14656 


4 73 


1.5539 


5.13 


1.6351 


3.94 


1.3712 


4.34 


1.4679 


j 4.74 


1.5560 


5.14 


16371 


3.95 


1.3737 


4.35 


1.4702 


4.75 


1 5581 


5.15 


1.6390 


3.96 


1.3762 


4.36 


1.4725 


4:76 


1 .5602 


5.16 


1.6409 


3 97 


1.3788 


4.37 


1.4748 


4.77 


1.5623 


5.17 


1.6429 


3.98 


1.3-13 


4.38 


1.4770 


4.78 


1.5644 


5.18 


1.6448 


3.99 


1 3838 


\ 4.39 


1.4793 


479 


1 56H5 


5.19 


1.6467 


4. 


1.3863 


| 4.40 


1.4816 


4.80 


1.6086 


5.20 


1.6487 



HYPERBOLIC LOGARITHMS. 



129 



N 


Log. 


N 


Log. 


N 


Log. 


N 


Log. 


5.21 


1 6506 


5.61 


1.7246 


6.01 


1.7934 


6.41 


1.8579 ' 


5.22 


1.6525 


5.62 


1.7263 


6.02 


1.7951 


6.42 


1.8594 


5 23 


1.6544 


5.63 


1.7281 


6.03 


1.7967 


6.43 


1.8610 


5.24 


1.6563 


5.64 


1.7299 


6.04 


1.7984 


6.44 


1.8625 


5.25 


1.6582 


5.65 


1.7317 


6.05 


1.8001 


6.45 


1.8611 


5.26 


1.6601 


5.66 


1.7334 


6.06 


1.8017 


046 


1.8656 


5.27 


1.6620 


5.67 


1.7352 


6.07 


1.8034 


6.47 


1.8672 


5.28 


1.6639 


5.68 


1.7370 


6.08 


1.8050 


6.48 


1.8687 


5.29 


1.6658 


5.69 


1.7387 


6.09 


1.8066 


6.49 


1.8703 


5.30 


1.6677 


5.70 


1.7405 


6.10 


1.8083 


6.60 


1.8718 


5.31 


1.6696 


5.71 


1.7422 


6.11 


1.8099 


6.51 


1.8733 


5.32 


1.6715 


5.72 


1.7440 


6.12 


1.8116 


652 


1.8749 


5.33 


1 6734 


5.73 


1 7457 


6.13 


1.8132 


6.53 


1.8764 


5.34 


1.6752 


5.74 


1.7475 


6.14 


1.8148 


6.54 


1.8779 


5.35 


1.6771 


5.75 


1.7492 


6.15 


1.8165 


6.55 


1.8795 


5.36 


1.6790 


5.76 


1.7509 


6 16 


1.8181 


6.56 


1 8810 


5.37 


1.6808 


5.77 


1.7527 


6.17 


1.8197 


6.57 


1.8S25 


5.38 


1.6827 


5.78 


1.7544 


6.18 


1.8213 


6.58 


18810 


5.39 


1.6845 


5.79 


1.7561 


6.19 


1.8229 


6.59 


1.8856 


5.40 


1.6864 


5.80 


1.7579 


6.20 


1.8245 


6.60 


1.8871 


541 


1.6882 


5.81 


1.7596 


6.21 


1.8262 


6.61 


1.8886 


5.42 


1.6901 


5.82 


1.7613 


6.22 


1.8278 


6 62 


1.8901 


5.43 


1.6919 


5 83 


1.7630 


6.23 


1.8294 


6 63 


1.8916 


5.44 


1.6938 


5.84 


1.7647 


6.24 


1.8310 


6.64 


1.8931 


5.45 


1.6956 


5.85 


1.7664 


6.25 


1.8326 


6.65 


1.8946 


5.46 


1.6974 


5.86 


1.7681 


6.26 


1.8342 


6.66 


1.8961 


5.47 


1.6993 


5.87 


1.7699 


6.27 


1.8358 


6.67 


1.8976 


5.48 


1.7011 


5.88 


1.7716 


6.28 


18374 


6.68 


1.8991 


5.49 


1.7029 


5.89 


1.7733 


6.29 


1.8390 


6.69 


1.9006 


5.50 


1.7047 


5.90 


1.7750 


6.30 


1.8405 


6.70 


1.9021 


5.51 


1.7066 


5.91 


1.7766 


6.31 


1 8421 


6.71 


1.9036 


5.52 


1.7084 


5.92 


1.7783 


6.32 


1.8437 


6.72 


1.9051 


5.53 


1.7102 


5.93 


1 7800 


6.33 


1.8453 


6.73 


1.9C66 


5.54 


1.7T0 


5.94 


1.7817 


6.34 


1.8469 


6.74 


1.9(181 


5.55 


1.7138 


5.95 


1.7834 


6.35 


1.8485 


6.75 


1.9095 


5.56 


1.7156 


5.96 


1.7851 


6.36 


1.8500 


6.76 


1.9110 


5.57 


1.7174 


5 97 


1.7867 


6.37 


1.8516 


6.77 


1.9125 


5.58 


1.7192 


6.98 


1.7.S84 


6.38 


1 8532 


6.78 


1.9140 


5.59 


1.7210 


5.99 


1 7901 


6 39 


1.8547 


6.79 


1.9155 


6.60 


1.7228 


6. 


1.7918 


6.40 


1.8563 


6.80 


1.9169 



130 



HYPERBOLIC LOGARITHMS. 



N 


1 w. 


! - 


Log. 


N 


Log. 


N 


Log. 


6.81 


1.9184 


7.21 


1.9755 


7.61 


2.0295 


8.01 


2.0807 


6.82 


1.919S 


7.22 


1.97: 


7.62 


2.0308 


- 02 


2.0819 


6.83 


1.9213 


7.23 


1.9782 


7.6? 


\ 2.0321 


8.03 


2.0s 3 2 


6.84 


1.9228 


7.24 


1.9796 


7.64 


2.0334 


8.04 


2.0844 


6.S5 


1.9242 


7.25 


1.9810 


7.65 


2.0347 


8.05 


2.0S57 


6.86 


1.9257 


7,26 


1.9824 


7.66 


2.0360 


8.06 


2.0869 


6.87 


1.9272 


7.27 


1.9838 


7.67 


2 0373 


8.07 


2.0882 


6.88 


1.9286 


7 25 


1.9851 


1 7.68 


2.0*86 


S.f8 


2.0894 


6.89 


1.9301 


7.29 


1.9865 


7.69 


2.0399 


8.09 


2.0906 


6.90 


1.9315 


7.30 


1.9879 


7.70 


i 2.0412 


810 


2.0919 


6.91 


1.9330 


7.31 


1.9892 


7.71 


2 042-' 


811 


2.0931 


6.92 


1.9344 


7,32 


1.9906 


7.72 


2.0438 


8.12 


2.0943 


6.93 


1.9359 


7.33 


1.9920 


7.73 


2.0451 


8.13 


2.0956 


6.94 


1.9373 


7.34 


1.9933 


7.74 


2.0464 


8.14 


2.0968 


6.95 


1.9357 


7.35 


1.9947 


7,75 


2.0477 


8.15 


2.0980 


6.96 


1.9402 


7.36 


1.9961 


7.76 


2.0490 


8.]6 


2.0992 


B 97 


1.9416 


7.37 


1.9974 


_ _ _ 


2 0503 


5,17 


2.1005 


5.98 


1.9430 


7.38 


1.9988 


! 7.78 


2.0516 


8.18 


2.1017 


6.99 


1.9445 


7.39 


2.0001 


! 7.79 


2.0528 


8.19 


2.1029 


7. 


1.9459 


7.40 


2.0015 


7.80 


2.0541 


8.20 


2.1041 


7.01 


1.9473 


7.41 


2.0028 


7. SI 


2.0554 


8.21 


2 1054 


7.02 


1.9488 


7.42 


2.0042 


7.82 


2.0567 


8.22 


2.1066 


7.03 


1.9502 


7.43 


2.0055 


7.53 


' 2.0580 


8.23 


2.1078 


7.04 


1.9516 


7,44 


2.0069 


7.84 


2.0592 


8 24 


2.1090 


7.05 


1.9530 


7.45 


2.0082 


7.85 


2.0605 


8.25 


2.1102 


7.06 


1.9544 


7.46 


2.0096 


7.86 


2.0618 


8.26 


2.1114 


7.07 


1.9552 


7.47 


2.0109 


7.87 


2.0631 


8.27 


2.1126 


7.08 


1.9573 


7.48 


2.0122 


7.55 


2.0643 


8.28 


2.1138 


7.09 


1.9557 


7.49 


2.0136 


7.S9 


2.0656 


8.29 


2.1150 


7.10 


1.9601 


7.50 


2.0149 


7.90 


2.0669 


5 30 


2.1163 


7.11 


1.9615 


7.51 


2.0162 


7.91 


2.0681 


8.31 


2 1175 


7.12 


1.9629 


7.52 


2.0176 


7.92 


2.0694 


8.32 


2.1187 


7.13 


1.9643 


7.53 


2.0189 


7.93 


2.0707 


8.33 


2.1199 


7.14 


1.9657 


7.54 


2.0202 


7.94 


2.0719 


8.34 


2.1211 


7.15 


1.9671 


7.55 


2.0215 


7.95 


2.0732 


8.35 


2.1223 


7.16 


1.9685 | 


7.56 


2.0229 


7.96 


2.0744 


8,36 


2.1235 


7.17 


1.9699 


7.57 


2.0242 


" M 


2.0757 


8.37 


2.1247 


7.18 


1.9713 


7.58 


2.0255 


7.98 


2.0769 


8.38 


2.1258 


7.19 


1.9727 


7.59 


2.0268 


7.99 


2.0782 


B.39 


2.1270 


7.20 


1.9741 


7.60 


2.0281 


8. 


2.0794 


8.40 ' 


2.1282 



HYPERBOLIC LOGARITHMS. 



«3* 



N 


Log. 


N 


Log. 


N 


Log. 


N 


Log. 


8.41 


2.1294 


881 


2.1759 


9.21 


2.2203 


9.61 


2.2628 


8.42 


2.130H 


8.82 


2.1770 


9.22 


2.2214 


9.62 


2.2638 


8.43 


2.1318 


8 83 


2.1782 


9.23 


2.2225 


9.63 


2.2649 


8.44 


2.1330 


8.84 


2.1793 


9.24 


2.2235 


9.64 


2.2659 


8.45 


2.1342 


8.85 


2.1804 


9.25 


2.2246 


9.65 


2.2670 


8.46 


2.1353 


8.86 


2.1815 


9.26 


2.2257 


9.66 


2.2680 


8.47 


2.1365 


8.87 


2.1827 


9.27 


2.2268 


9.67 


2.2690 


8.48 


2.1377 


8.88 


2.1838 


9.28 


2.2279 


9.68 


2.2701 


8.49 


2.1389 


8.89 


2.1849 


9.29 


2.2289 


9.69 


2.2711 


8.50 


2.1401 


8.90 


2.1861 


9.30 


2.2300 


9.70 


2.2721 


8.51 


2.1412 


8.91 


2.1872 


9.31 


2.2311 


9.71 


2.2732 


8.52 


2.1424 


8.92 


2.1883 


9.32 


2.2322 


9.72 


2.2742 


8.53 


2.1436 


8.93 


2.1894 


9.33 


2.2332 


9.73 


2.275*2 


8.54 


2.1448 


8.94 


2.1905 


9.34 


2.2343 


9.74 


2.2762 


8.55 


2.1459 


8.95 


2.1917 


9.35 


2.2354 


9.75 


2.2773 


8.56 


2.1471 


8.96 


2.1928 


9.36 


2.2364 


9.76 


2.2783 


8.57 


2.1483 


8.97 


2.1939 


9.37 


2.2375 


9.77 


2.2793 


8.58 


2.1494 


8.98 


2.1950 


9.38 


2.2386 


9.78 


2.2803 


8.59 


2.1506 


8.99 


2.1961 


9.39 


2.2396 


9.79 


2.2814 


8.60 


2.1518 


9. 


2.1972 


9.40 


2.2407 


9.80 


2.2824 


8.61 


2.1529 


9.01 


2.1983 


9.41 


2.2418 


9.81 


2.2834 


8.62 


2.1541 


9.02 


2.1994 


9.42 


2.2428 


9.82 


2.2844 


8.63 


2.1552 


9.03 


2.2006 


9.43 


2.2439 


9.83 


2.2854 


8.64 


2.1564 


9.04 


2.2017 


9.44 


2.2450 


9.84 


2.2865 


8.65 


2.1576 


9.05 


2.2028 


9.45 


2.2460 


9.85 


2.2875 


8.66 


2.1587 


9.06 


2.2039 


9.46 


2.2471 


9.86 


2.2885 


8.67 


2.1599 


9.07 


2.2050 


9.47 


2.2481 


9.87 


2.2895 


8.68 


21610 


9.08 


2.2061 


9.48 


2.2492 


9.88 


2.2905 


8.69 


2.1622 


9.09 


2.2072 


9.49 


2.2502 


9.89 


2.2915 


8.70 


2.1633 


9.10 


2.2083 


9.50 


2.2513 


9.90 


2.2925 


8.71 


2.1645 


9.11 


2.2094 


9.51 


2.2523 


9.91 


2.2935 


8.72 


2.1656 


9.12 


2.2105 


9.52 


2.2534 


9.92 


2.2946 


8.73 


2.1668 


9.13 


2.2116 


9.53 


2.2544 


9.93 


2.2956 


8.74 


2.1679 


9.14 


2.2127 


9.54 


2.2555 


9.94 


2.2966 


8.75 


2.1691 


9.15 


2.2138 


9.55 


2.2565 


9.95 


2.2976 


8.76 


2.1702 


9.16 


2.2148 


9.56 


2.2576 


9.96 


2 2986 


8.77 


2.1713 


9.17 


2.2159 


9.57 


2.2586 


9.97 


2.2996 


8.78 


2.1725 


9.18 


2.2170 


9.58 


2.2597 


9.98 


2.3006 


8.79 


2.1736 


9.19 


2.2181 


9.59 


2.2607 


9.99 


2.3016 


8.80 


2.1748 


9.20 


2.2192 


9.60 


2.2618 


10. 


2.3026 



132 



HYPERBOLIC LOGARITHMS. 



N 


Log. 


N 


Log. 


N 


Log. 


N 


Log. 


10.1 


2.3126 


14.1 


2.6462 


18.1 


2.8959 


22.1 


3.0956 


10.2 


2.3225 


14.2 


2.6532 


18.2 


2.9014 


22 2 


3 1001 


10.3 


2.3322 


14.3 


2.6602 


18 3 


2 9069 


22.3 


3.1046 


10.4 


2.3419 


14.4 


2.6672 


18.4 


2.9123 


22.4 


3.1090 


10.5 


2.3515 


14.5 


2.6741 


18.5 


2.9178 


22.5 


3.1135 


10.6 


2.3609 


14.6 


2.6810 


18.6 


2 9231 


22 6 


3.1179 


10.7 


2 3703 


14.7 


2.6878 


18.7 


2 9285 


22.7 


3.1224 


10.8 


23796 


14.8 


2.6946 


18.8 


2.9338 


22.8 


3.1267 


10.9 


2 3888 


14.9 


2.7013 


18.9 


2.9391 


22.9 


3.1311 


11 


2.3979 


15. 


2.7080 


19. 


2.9444 


23. 


3.1355 


11.1 


2.4070 


15.1 


2 7147 


19.1 


2.9497 


23.1 


3.1398 


11.2 


2.4160 


15.2 


2.7213 


19.2 


2.9549 


23 2 


3.1441 


11.3 


2.4249 


15.3 


2.6279 


X 9.3 


2 9601 


23.3 


3 1484 


11.4 


2.4337 


15.4 


2.7344 


19.4 


2.9653 


23.4 


3 1527 


11.5 


2.4424 


15.5 


2.7408 


19.5 


2.9704 


23.5 


3.1570 


11.6 


2.4510 


15.6 


2.7472 


19.6 


2.9755 


23.6 


3.1612 


11.7 


2 46U6 


15.7 


2.7536 


19.7 


2.9806 


23.7 


3.1655 


11.8 


2 4681 


158 


2.7600 


198 


2.9856 


23.8 


3.1697 


11.9 


2.4765 


15.9 


2.7663 


19.9 


2.9907 


23.9 


3.1739 


12 


2.4849 


16. 


2.7726 


20. 


2.9957 


24. 


3.1780 


12.1 


2.4932 


16.1 


2.7788 


20.1 


3 0007 


24.1 


3.1822 


12.2 


2.5014 


16.2 


2.7850 


20.2 


3.0057 


24.2 


3.1863 


12.3 


2.5096 


16.3 


2.7912 


20.3 


3.0106 


24.3 


3.1905 


12.4 


2.5178 


16.4 


2.7973 


20.4 


3.0155 


24.4 


3 1946 


12.5 


2.5259 


16.5 


2.8033 


20.5 


3.0204 


24.5 


3.1987 


12.6 


2.5338 


16.6 


2.8094 


20.6 


3.0253 


24.6 


3.2027 


12.7 


25417 


16.7 


2.8154 


20.7 


3.0301 


24.7 


3.2068 


12.8 


2.5495 


16.8 


2.8214 


20.8 


3.0349 


24.8 


3.2108 


12.9 


2.5572 


16.9 


2 8273 


20.9 


3.0397 


24.9 


3.2149 


13 


2.5649 


17. 


2.8332 


21. 


3.0445 


25. 


3.2189 


13.1 


2.5726 


17.1 


2.8391 


21.1 


3.0493 


25.5 


3.2387 


13.2 


2.5802 


17.2 


2.84i 9 


21.2 


3.0540 


26. 


3.2581 


13 3 


2.5877 


17.3 


2.8507 


21.3 


3.0587 


26.5 


3.2771 


13.4 


2.5952 


17.4 


2.8565 


21.4 


3.0634 


27. 


3.2958 


13.5 


2.6027 


17.5 


2.8622 


21.5 


3.0680 


27.5 


3.3142 


13.6 


2.6101 


17.6 


2.8679 


21.6 


3.0727 


28. 


3.3322 


13.7 


2.6174 


17.7 


2.8735 


21.7 


3.0773 


28.5 


3.3499 


13.8 


2.6247 


17.8 


2.8792 


21.8 


3.0819 


29. 


3 3673 


13.9 


2.6319 


17.9 


2 8848 


21 9 


3.0*66 


29.5 


3.3844 


14. 


2.6391 


18. 


2.8904 


22. 


3.0910 


30. 


3.4012 



Weights anb Measures. 



The yard is the standard unit for length in the United 
States and Great Britain. To determine the length of the yard, 
a pendulum vibrating seconds in a vacuum at the level of the 
sea in the latitude of London, with the Fahrenheit thermometer 
at 62°, is supposed to be divided into 391,393 equal parts ; 360,- 
000 of these parts is the length of the standard yard. Actually, 
the standard yard in both the United States and Great Britain 
is a metallic scale made with great care and kept by the respec- 
tive governments, and from this standard other measures of 
length have Deen produced. 

The standard unit of weight in the United States and 
Great Britain is the Troy pound, which is equal in weight to 
22.2157 cubic inches of distilled water at 62° Fahrenheit, the 
barometer being 30 inches. The Troy pound contains 5,760 
Troy grains; the Pound Avoirdupois, which is the unit of 
weight used in commercial transactions and mechanical cal- 
culations in the United States and Great Britain, is equal to 
7,000 Troy grains. 

In the United States the standard unit of liquid measure 
is the wine gallon, containing 231 cubic inches or 8.3389 pounds 
avoirdupois of distilled water at a temperature of its greatest 
density (39 o -40° F). 

In the United States the standard unit for dry measure is 
the Winchester Bushel, containing 2150.42 cubic inches. 

In Great Britain the standard measure for both liquid and 
dry substances is the Imperial Gallon, which is denned as the 
volume of 10 pounds avoirdupois of distilled water, when weighed 
at 62° Fahrenheit with the barometer at 30 inches. The Im- 
perial Gallon contains 277.463 cubic inches. The Imperial 
Bushel of 8 gallons contains 2219.704 cubic inches. 

Long Measure. 

12 inches = 1 foot = 0.30479 meters. 

3 feet = 1 yard = 0.91437 meters. 

h%. yards = 1 rod or pole = 16% feet = 198 inches. 

40 rods = 1 furlong = 220 yards = 660 feet. 

8 furlongs = 1 statute or land mile = 320 rods = 1760 yards. 

3 miles = 1 league = 24 furlongs = 960 rods. 

5280 feet = 1 statute or land mile = 1.609 kilometer. 

1 geographical or nautical mile = 1 minute = -^ degree. 

(i33) 



134 WEIGHTS AND MEASURES. 

As adopted by the British admiralty,* a nautical mile is 6080 ft. 
1 nautical mile = 1.1515 statute or land miles. 
1 statute or land mile = 0.869 nautical miles. 

Square Measure. 

1 square yard = 9 square feet = 0.836 square meters. 
1 square foot = 144 square inches = 929 square centi- 
meters. 

1 square inch = 6.4514 square centimeters. 
A section of land is 1 mile square = 640 acres. 
1 acre = 43,560 square feet = 0.40467 hectare. 
1 square acre is 208.71 feet on each side. 

Cubic ileasure. 

1 cubic yard = 27 cubic feet = 0.7645 cubic meters. 
1 cubic yard = 201.97 (wine) gallons = 7.645 hectoliter. 
1 cubic foot = 1728 cubic inches = 28315.3 cubic centi- 
meters. 

1 cubic foot = 7.4805 (wine) gallons = 28.315 liters. 

Note. — 1 cubic foot contains 6.2355 imperial (English) gal- 
lons. 

A cord of wood = 128 cubic feet, being 4X4X8 feet. 
A perch of stone = 24% cubic feet, being 16>^ X 1% X 1 
foot, but it is generally taken as 25 cubic feet. 

Liquid Hea&ure. 

1 pint = 28.88 cubic inches. 

2 pints = 1 quart = 57.75 cubic inches = 0.9463 liter. 

4 quarts = 1 gallon = 231 cubic inches = 3.7852 liters. 
Note. — 1 imperial (English) gallon is 277.274 cubic inches. 

Dry Measure. 

1 standard U. S. bushel = 2150.42 cubic inches. 

1 standard U. S. bushel = 4 pecks. 

1 peck = 2 gallons = 8 quarts. 

1 gallon = 4 quarts = 268* cubic inches. 

1 quart = 2 pints = 67A cubic inches. 

100 bushels (approximately) = 124 >£ cubic feet. 

80 bushels (approximately) = 100 cubic feet. 

Avoirdupois Weight. 

(Used in business and mechanical calculations.) 

1 pound = 16 ounces = 0.45359 kilograms. 

1 ton = 2240 pounds. A short ton is 2000 pounds. 

* See Machinery, page 23, Sept., 1897. 



WEIGHTS AND MEASURES. 



135 



Troy Weight. 

(Used when weighing gold, silver and jewelry.) 

1 pound = 12 ounces = 0.37324 kilogram. 

' ^hts = 1.0971 ounces avdp. 



x puuiiu ±4 UUULCS 

1 ounce = 20 pennyweig 

Apothecaries' Weight. 



1 pound = 1 pound troy weight 
1 ounce = 8 drachms. 
1 drachm = 3 scruples. 
1 scruple = 20 grains. 



12 ounces. 



Weights of Produce. 

The following are the weights of certain articles of produce: 



Pounds 


Pc 


unds 


Pounds 


per bi 


ashel. 


per 


bushel. 


per bushel. 


Wheat, 


60 


Oats, 


32 


White potatoes, 60 


Corn in the ear, 


70 


Peas, 


60 


Sweet potatoes, 55 


Corn shelled, 


56 


Ground peas, 


24 


Onions, 57 


Rye, 


56 


Corn meal, 


48 


Turnips, 57 


Buckwheat, 


48 


Malt, 


38 


Clover seed, 60 


Barley, 


48 


White beans, 


60 


Timothy seed, 45 



The rietric System of Weights and Heasures. 

The unit in the metric system is the meter. The length of 
the meter was intended to be toooWoo ( one ten-millionth part) 
of the length of a quadrant of the earth through Paris, which 
is the same as tfooVooo °^ an y quadrant from either pole to 
equator. 

By later calculations it has been ascertained that the meter 
as first adopted and now used is slightly too short according to 
this theoretical requirement, but this, of course, makes no dif- 
ference ; because, practically speaking, the length of a meter is 
the length of a certain standard meter kept at Paris in the care 
of the French government, and it is from this standard meter 
(and not from the quadrant of the earth) that all other standard 
meters kept for reference are derived. 

The gram, which is the unit of weight, is the weight of 1 
cubic centimeter of water at its maximum density, which is at 
4° C. (39° to 40° F.) 

The commercial denomination used for weight is the kilo- 
gram = 1000 grams = 1 cubic decimeter = 1 liter of water at 
maximum density. 

The metric system has been adopted by Mexico, Brazil, 
Chile and Peru, and by all European countries except England 
and Russia. 



136 



WEIGHTS AND MEASURES. 



Length. 

1 Meter = 10 Decimeters 

1 Decimeter = 10 Centimeters 
1 Centimeter = 10 Millimeters 
1 Millimeter = —Vo Meter 
1 Decameter = 10 Meters 
1 Hektometer= 100 Meters 
1 Kilometer = 1000 Meters 
1 Myriameter = 10000 Meters 



= 39.37 inches. 
= 3.937 inches. 
= 0.3937 inches. 
== 0.03937 inches. 

= 32 feet 9.7 inches. 
= 328 " 1 inch. 
= 0.6214 mile. 
= 6.214 miles. 






Area. 

1 square millimeter = 0.00155 square inch. 
100 square millimeters = 1 square centimeter = 0.155 sq. inch. 

= 15.5 sq. inch. 
= 10.764 sq. feet. 
= 1550 sq. inches. 
= 119.6 sq. yards. 
= 2.471 acres. 



100 " centimeters = 1 " decimeter 

100 " decimeters =1 " meter 

1 Centare = 1 " meter 

1 Are = 100 " meters 



1 Hectare 



10000 sq. meters 



Solids. 

1 cubic millimeter = 0.000061 cubic inch. 
1000 cubic millimeters = 1 cubic centimeter = 0.061 cubic inch. 
1000 " centimeters = 1 " decimeter =61.027 " " 
1000 " decimeters = 1 " meter = 35.3 " feet. 



1 liter 
1 deciliter 
1 centiliter 
1 milliliter 
1 decaliter 
1 hectoliter 
1 kiloliter 
1 liter = 61.0 



Liquid. 

10 deciliters 
10 centiliters = 
10 milliliters = 
T7) oir liters 
10 liters 
100 liters 
1000 liters 
27 cubic inches 



1 cubic decimeter. 
100 cubic centimeter. 
10 cubic centimeters. 
1 cubic centimeter. 
10 cubic decimeters. 
100 cubic decimeters. 
1 cubic meter. 
1.0567 quarts. 



WEIGHTS AND MEASURES. 1 37 

TABLE No. 7. Reducing Millimeters to Inches. 



Mm. 


Inches. 


Mm. 


Inches. 


Mm. 


Inches. 


0.02 


0.00079 


0.52 


0.02047 


2 


0.07874 


0.04 


0.00157 


0.54 


0.02126 


O 


0.11811 


0.06 


0.00236 


0.56 


0.02205 


4 


0.15748 


0.08 


0.00315 


0.58 


0.02283 


5 


0.19685 


0.10 


0.00394 


0.60 


0.02362 


6 


0.23622 


0.12 


0.00472 


0.62 


0.02441 


7 


0.27559 


0.14 


0.00551 


0.64 


0.02520 


8 


0.31496 


0.16 


0.00630 


0.66 


0.02598 


9 


0.35433 


0.18 


0.00709 


0.68 


0.02677 


10 


0.39370 


0.20 


0.00787 


0.70 


0.02756 


11 


0.43307 


0.22 


0.00866 


0.72 


0.02835 


12 


0.47244 


0.24 


0.00945 


0.74 


0.02913 


13 


0.51181 


0.26 


0.01024 


0.76 


0.02992 


14 


0.55118 


0.28 


0.01102 


0.78 


0.03071 


15 


0.59055 


0.30 


0.01181 


0.80 


0.03150 


16 


0.62992 


0.32 


0.01260 


0.82 


0.03228 


17 


0.66929 


0.34 


0.01339 


0.84 


0.03307 


18 


0.70866 


0.36 


0.01417 


0.86 


0.03386 


19 


0.74803 


0.38 


0.01496 


0.88 


0.03465 


20 


0.78740 


0.40 


0.01575 


0.90 


0.03543 


21 


0.82677 


0.42 


0.01654 


0.92 


0.03622 


22 


0.86614 


0.44 


0.01732 


0.94 


0.03701 


23 


0.90551 


0.46 


0.01811 


0.96 


0.03780 


24 


0.94488 


0.48 


0.01890 


0.98 


0.03858 


25 


0.98425 


0.50 


0.01969 


1.00 


0.03937 


26 


1.02362 


TABLE No. 8. 


Reducing Inches 1 


.0 Hi Hi meters. 


Inches. 


Mm. 


Inches. 


Mm. 


Inches. 


Mm. 


1 

1 6 


1.59 


1 3 

16 


20.64 


2% 


57.15 


X 


3.17 


7 A 


22.22 


%y 2 


63.50 


3 
16 


4.76 


1 5 
16 


23.81 


3 


76.20 


X 


6.35 


1 


25.40 


4 


101.6 


5 
1 6 


7.94 


1H 


28.57 


5 


127 


H 


9.52 


IX 


31.75 


6 


152.4 


TTT 


11.11 


1H 


34.92 


7 


177.8 


% 


12.70 


iy* 


38.10 


8 


203.2 


9 

1 6 


14.29 


iy* 


41.27 


9 


228.6 


H 


15.87 


ltf 


44.45 


10 


254 


1 1 

1 6 


17.46 


i% 


47.62 


11 


279.4 


M 10.05 


2 


50.80 


12 


304.8 



13° WEIGHTS AND MEASURES. 

Table of Reduction for Pressure per Unit of Surface. 

1 kilogram per sq. centimeter = 14.223 pounds per sq. inch. 
1 kilogram per sq. centimeter = 0.968 atmosphere. 
1 pound per sq. inch = 0.0703 kilograms per sq. centimeter. 
1 pound per sq. inch = 0.06S atmosphere. 

Table of Reduction for Length and Weight. 

1 kilogram per kilometer = 3.54S pounds per mile. 

1 kilogram per meter = 0.672 pounds per foot. 

1 pound per mile = 0.282 kilograms per kilometer. 

1 pound per foot = 1.4S8 kilograms per meter. 

Weight of Water 4 C.) 

1 cubic cm. weighs 1 gram. 

1 cubic inch weighs 0.036125 pounds = 16.386 grams. 
1 liter weighs 1 kilogram = 2.2046 pounds. 

1 quart weighs 2.0862 pounds = 0.9463 kilograms. 

1 cubic meter weighs 1000 kilograms = 2204.6 pounds. 
1 cubic foot weighs 62.425 pounds = 28.32 kilograms. 

Measure of Water. 

1 kilogram measures 1 liter =1.057 quarts. 

1 kilogram measures 0.353 cubic feet = 61.03 cubic inches. 

1 pound measures 0.01602 cubic ft. = 0.454 liter. 

1 pound measures 27.68 cubic ins. = 453.59 cubic centimeters. 



SPECIFIC GRAVITY. 

The specinc gravity of a body is its weight as compared 
with the weight of an equal volume of another body which is 
adopted as a standard. For all solid substances, water at its 
maximum density (4 s C.) is the usual standard. For instance, 
the specific gravity of zinc is 7 : this simply means that one 
cubic foot of zinc is 7 times as heavy as one cubic foot of 
water. One cubic foot of water weighs 62.425 pounds. There- 
fore, by multiplying the specific gravity of any solid body by 
62.425 its weight per cubic foot is obtained. In the metric 
system of measure and weight one cubic centimeter of water 
weighs one gram; therefore the table of specific gravity will 
also directly give the weight of the material in grams per cubic 
centimeter, in kilograms per cubic decimeter, or in 1000 kilo- 
grams (the so-called metric ton] per cubic meter, 



WEIGHTS AND MEASURES. 



139 



TABLE No. 9. 


Specific Gravity, 


Weig 


hts 


and Values. 




Metric. 


American. 


Approximate 


Metals. 


Kilog. per 






value 
per 




cubic dec. c 


r Pounds per 


Pounds per 


pound. 




specific 


cubic inch. 


cubic foot. 




gravity. 








Water .... 


1 


0.036125 


62.425 




Gold (24 k) . . 


19.361 


0.697 


1208 


$299.70 


Platinum . . . 


21.531 


0.775 


1344 


122.00 


Silver .... 


10.474 


0.377 


654 


12.14 


Wrought iron . 


7.78 


0.28 


485 


0.015 


Cast iron . . . 


7.21 


0.26 


450 


0.008 


Tool Steel . . 


7.85 


0.284 


490 


0.10 


Zinc 


7 


0.252 


437 


0.10 


Antimony . . . 


6.72 


0.242 


419 


0.12 


Copper .... 


8.607 


0.31 


537 


0.15 


Mercury .... 


13.596 


0.489 


849 




Tin . . . 


7.291 


0.262 


455 


0.25 


Aluminum . . 


2.67 


0.096 


166 




Lead 


11.36 


0.408 


708 


0.05 


TABLE No. 10. 


Specific Gravity and Weight of Medium 




Dry Wood. 


Variety. 




Metric. 


American. 












Kilog. per cubic 


Pounds per cubic 






dec. specific gravity. 


foot. 


Birch 




0.60 to 0.80 


37.5 to 50 


Ash 




0.50 to 0.80 


31 to 50 






0.60 to 0.80 


37.5 to 50 


Oak 




0.60 to 0.90 


37.5 to 56 






1.19 


74 


Lignum-vitae 




1.33 


83 


Spanish mahoga 


ny . . . 


0.85 


53 






0.50 


32 






0.50 


32 


Pine , . , , , 


• ti« 


0,40 to 0,80 


25 to 50 


Pitch pine. , , , 


JL.„-U»-V-~— - — 


0,80 

J- 1 ■ !"" 


50 



140 



WEIGHTS AND MEASURES. 



TABLE No. 11. Specific Gravity and Weight per Cubic 
Foot of Various flaterials. 

(The weight may vary according to the properties of the material). 



Materials. 



Asphalt .... 

Brick 

Gray granite . . 
Red granite . . 
Limestone . . . 

Sand 

Portland cement 
Brickwork . . 

Slate 

Glass 

Emery .... 
Grindstone . . 

Coal 

Porcelain . . . 
Lime 



Metric. 



Kilog. per cubic dec. 
specific gravity. 



to 



1.4 

1.6 to 

2.4 

2.5 

2.7 

1.5 

1.26 

1.75 

2.8 

2.52 

4.0 

2.4 

1.5 

2.4 

0.96 



American. 



Pounds 
per cubic foot. 



87 
100 to 125 

150 

157 to 187 

168 

94 

78 
110 
175 
157 
250 
150 

94 
150 

60 



TABLE No. 12. Specific Gravity and Weight of Liquids. 





>> 


Metric. 


American. 


Liquids. 












Kilog. 


Kilog. 


Pounds 


Pounds 




per cubic 


per 


per 


per 






dec. 


liter. 


cub. inch. 


gallon. 


Water .... 


1 


1 


1 


0.036125 


8.33 


Sea water . . 


1.03 


1.03 


1.03 


0.037 


8.55 


Sulphuric acid . 


1.841 


1.841 


1.841 


0.067 


15.48 


Muriatic acid . 


1.2 


1.2 


1.2 


0.043 


9.93 


Nitric acid . . 


1.217 


1.217 


1.217 


0.044 


10.16 


Alcohol .... 


0.833 


0.833 


0.833 


0.03 


6.93 


Linseed oil . . 


0.94 


0.94 


0.94 


0.034 


7.85 


Turpentine . . 


0.87 


0.87 


0.87 


0.031 


7.16 


Petroleum . . 


0.878 


0.878 


0.878 


0.032 


7.39 


Machine oil . . 


0.9 


0.9 


0.9 1 


0.0324 


7.5 



WEIGHTS AND MEASURES. 141 

To Calculate Weight of Casting from Weight of Pattern. 

When pattern is made from pine and no nails used, the 
rule is : Multiply the weight of the pattern by 17 and the prod- 
uct is the weight of the castings. 

When nails are used in the pattern, multiply its weight by 
a little less, probably 15 or 16. 

When the pattern has core prints, their weight must be 
calculated and also the weight of what there is to be cored out 
in the casting, which must all be deducted. This mode of cal- 
culating the weight of castings is, of course, only approxima- 
tion, but it is frequently very useful. 



Weight of an Iron Bar of any Shape of Cross Section. 

A wrought iron bar of 1 square inch area of cross section 
and one yard long weighs 10 pounds. Therefore, the weight of 
wrought iron bars of any shape, as, for instance, railroad rails, 
I beams, etc., may very conveniently be obtained by first mak- 
ing a correct, full size drawing of the cross section and measur- 
ing its area by a planimeter, which gives the area in square 
inches. Multiply this area by 10 and the product is the weight 
in pounds per yard; or multiply the area by 3.33 and the product 
is the weight in pounds per foot. 



To Calculate Weight of Sheet Iron of any Thickness. 

One square foot of wrought iron, 1 inch thick, weighs very 
nearly 40 pounds (40.2 pounds) and one square foot ¥ y, which 
is T ff o thick, weighs 1 pound. Therefore, a practical rule for 
quick calculation of the weight of sheet iron is : Divide the 
thickness of the iron as measured by a micrometer calliper in 
thousandths of inches by 25, and the quotient is the weight in 
pounds per square foot. 



To Calculate the Weight of Metals Not Given in the Tables. 

Find the weight of wrought iron, and multiply by the fol- 
lowing constants : 

Weight of wrought iron X 0.928 = cast iron. 

" X 1.014 = steel. 

" " " " X 0.918 = zinc. 

" " " " X 1.144 = copper. 

«* « « «« X 1.468 = lead. 



142 



WEIGHTS AND MEASURES. 



To Calculate the Weight of Zinc, Copper, Lead, etc., 

in Sheets. 

First find the weight by the rule given for sheet iron, and 
multiply by the constant as given in the above table, and the 
product is the weight of each metal in pounds per square foot. 

To Calculate the Weight of Cast Iron Balls. 

Multiply the cube of the diameter in inches by 0.1377, and 
the product is the weight of the ball in pounds. 

Thus : 

W—D^y, 0.1377. D = 1.936 a/W 

D = diameter of ball in inches. 

IV = weight of ball in pounds. 

In metric measure, multiply the cube of the diameter in 
centimeters by 0.003775, and the product is the weight of the 
ball in kilograms. 

Thus : 

W — M s X 0.003775. M — 6.422 X s/W 

W = weight in kilograms. 

M — diameter of ball in centimeters. 



TABLE No. 13. Weight of Round Steel per Lineal Foot. 

Steel weighing 489 pounds per Cubic Foot. 



Diameter in 


Weight 


Diameter in 


Weight Per 


Diameter in 


Weight Per 


inches. 


Per Foot. 


Inches. 


Foot. 


Inches. 


Foot. 


1 

1 6 


.0104 


1A 


3.011 


2 % 


12.044 


% 


.042 


•A 


3.375 


% 


13.503 


3 
1 6 


.091 


3 
16 


3.761 


H 


15.045 


% 


.167 


X 


4.168 


% 


16.67 


5 
16 


.261 


5 
16 


4.595 


Y% 


18.379 


H 


.375 


H 


5.043 


y* 


20.171 


7 
1 6 


.511 


T6 


5.512 


7 A 


22.047 


A 


.667 


*A 


6.001 


3 


24.005 


9 

1 6 


.844 


9 
T~6 


6.512 


A 


26.048 


A 


1.042 


H 


7.043 


% 


28.173 


11 

1 6 


1.261 


1 1 

16 


7.596 


H 


30.382 


u 


1.5 


H 


8.169 


A 


32.674 


13 

1 6 


1.761 


1 3 

1 6 


8.702 


A 


35.05 


A 


2.042 


7 A 


9.377 


% 


37. SOS 


15 

1 6 


2.344 


1 6 


10.013 


A 


40.05 


1 


2.667 


2 


10.669 


4 


42.675 



WEIGHTS AND MEASURES. 



143 



TABLE No. 1 4. Weights of Square and Round Bars of 
Wrought Iron in Pounds per Lineal Foot. 

(Iron weighing 480 pounds per cubic foot). 



Thickness 


Weight of 


Weight of 


Thickness 


Weight of 


Weight of 


or Diameter 


Square Bar 


Round Bar 


or Diameter 


Square Bar 


Round Bar 


in 


One Foot 


One Foot 


in 


One Foot 


One Foot 


Inches. 


Long. 


Long. 


Inches. 


Long. 


Long. 


Vl6 


.013 


.010 


2 YlG 


21.89 


17.19 


Vs 


.052 


.041 


% 


22.97 


18.04 


YlG 


.117 


.092 


n /l6 


24.08 


18.91 


X 


.208 


.164 


% 


25.21 


19.80 


YlG 


.320 


.256 


1:i /ie 


26.37 


20.71 


3/ 
/& 


.469 


.368 


X 


27.55 


21.64 


Vie 


.638 


.501 


X %6 


28.76 


22.59 


1/ 

X2 


.833 


.654 


3 


30 


23. 56 


YlG 


1.055 


.828 


^16 


31.26 


24.55 


% 


1.302 


1.023 


% 


32.55 


25.57 


U /l6 


1.576 


1.237 


YlG 


33.87 


26.60 


3/ 
/4 


1.875 


1.473 


X 


35.21 


27.65 


»%6 


2.201 


1.728 


YlG 


36.58 


28.73 


% 


2.551 


2.004 


/8 


37.97 


29.82 


1 %6 


2.930 


2.301 


YlG 


39.39 


30.94 


1 


3.333 


2.618 


v 

/2 


40.83 


32.07 


YlG 


3.763 


2.955 


YlG 


42.30 


33.23 


% 


4.219 


3.313 


% 


43.80 


34.40 


•Tie 


4.701 


3.692 


ll /w 


45.33 


35.60 


1/ 

/4 


5.208 


4.091 


3/ 

/4 


46.88 


36.82 


•YlG 


5.742 


4.510 


X %6 


48.45 


38.05 


% 


6.302 


4.950 


7/ 


50.05 


39.31 


YlG 


6.888 


5.410 


!YlG 


51.68 


40.59 


/2 


7.5 


5.890 


4 


53.33 


41.89 


YlG 


8.138 


6.392 


YlG 


55.01 


43.21 


% 


8.802 


6.913 


1/ 
/a 


56.72 


44. 55 


"Ae 


9.492 


7.455 


YlG 


58.45 


45.91 


3/ 
/4 


10.21 


8.018 


1/ 

/4 


60.21 


47.29 


HAg 


10.95 


8.601 


YlG 


61.99 


48.69 


V 


11.72 


9.204 


3/ 
/8 


63.80 


50.11 


Wie 


12.51 


9.828 


YlG 


65.64 


51.55 


2 


13.33 


10.47 


1/ 

/2 


07.50 


53.01 


YlG 


14.18 


11.14 


YlG 


69.39 


54.50 


/8 


15.05 


11.82 


% 


71.30 


56 


YlG 


15.95 


12.53 


ll /l6 


73.24 


57.52 


1/ 

/4 


16.88 


13.25 


/4 


75.21 


59.07 


YlG 


17.83 


14 


*%6 


77.20 


60.63 


3 / 

/8 


18.80 


14.77 


7/ 
/8 


79.22 


62.22 


YlG 


19.80 


15.55 


l YlG 


81.26 


63.82 


X 


20.83 


16.36 


5 


83.33 


65.45 



144 



WEIGHTS AND MEASURES. 



TABLE No. i4. — (Continued). 



Thickness 


Weight of 


Weight of 


Thickness 


Weight of 


I 

Weight of 


or Diameter 


Square Bar 


Round Bar 


or Diameter 


Square Bar 


Round Bar 


in 


One Foot 


One Fool 


in 


One Foot 


One Foot 


Inches. 


Long. 


Long. 


Inches. 


Long. 


Long. 


5 M.6 


85.43 


67.10 


7% 


169.2 


132.9 


X 


87.55 


68.76 


X 


175.2 


137.6 


%6 


89.70 


70.45 


% 


181.3 


142.4 


1/ 


91.88 


72.16 


X 


1S7.5 


147.3 


%6 


94.08 


73.89 


% 


193.S 


152.2 


/8 


96.30 


75.64 


X 


200.2 


157.2 


%6 


98.55 


77.40 


X 


206.7 


162.4 


1/ 
/2 


100.8 


79.19 


s 


213.3 


167.6 


%6 


103.1 


81.00 


x 


226.9 


178.2 


/8 


105.5 


82.83 


x 


240.8 


189.2 


Hie 


107.8 


84.69 


x 


255.2 


200.4 


V 

/A 


110.2 


86.56 


9 


270.0 


212.1 


!%6 


112.6 


88.45 


/4 


285.2 


224.0 


X 


115.1 


90.36 


X 


300.8 


236.3 


15 /ie 


117.5 


92.29 ! 


X 


316.9 


248.9 


6 


120.0 


94.25 


10 


333.3 


261.8 


% 


125.1 


98.22 ! 


3^ 


350.2 


275.1 


% 


130.2 


192.3 


X 


367.5 


288.6 


% 


135.5 


106.4 


% 


385.2 


302.5 


i/ 

/2 


140.8 


110.6 


n 


403.3 


316.8 


V 


146.3 


114.9 


^ 


421.9 


331.3 


/4 


151.9 


119.3 


X 


440.8 


346.2 


•a 


157.6 


123.7 


% 


460.2 


361.4 


7 


163.3 


128.3 


12 


480. 


377. 



TABLE No. 15. Weight of Flat Iron in Pounds per Foot. 



Inches Vie \Y S %e % %e % 7 Ae % % % % 



1% 
IX 
1% 

IX 

IX 
IX 

2 



0.11 
0.13 
0.16 
0.18 
0.21 
0.24 
0.26 
0.29 
0.32 
0.34 
0.37 
0.40 
0.42 



0.21 
0.26 
0.32 
0.37 
0.42 
0.47 
0.53 
0.58 
0.63 
0.68 
0.74 
0.79 
0.84 



0.32 
0.40 
0.47 
0.55 
0.63 
0.71! 
0.79 
0.87 
0.95 
1.03 
1.11 
1.18 
1.26 



0.42 
0.53 
0.63 
0.74 
0.84 
0.95 
1.05 
1.16 
1.26 
1.37 
1.47 
1.58 
1.6S 



0.53 
0.66 
0.79 
0.92 
1.05 
1.18 
1.32! 
1.45 
1.58 
1.71 
1.84 
1.97 
2.11 



0.63 
0.79 
0.95 
1.11 
1.26 
1.42 
1.58 
1.74 
1.90 
2.05 
2.21 
2.37 
2.53 



0.73 


0.84 


0.92 


1.06 


1.11 


1.26 


1.29 


1.48 


1.47 


1.68 


1.66 


1.90 


1.84 


2.11 


2.03 


2.32 


2.21 


2.53 


2.39 


2.74 


2.58 


2 95 


2.76 


3.16 


2.95 


3.37 



1.31 
1.58 

1.85 
2.11 
2.37 
2.63 
2.89 
3.16 
3.42 
3.68 
3.95 
4.21 



1.90 
2.22 
2.53 
2.84 
3.16 
3.47 
3.79 
4.11 
4.42 
4.74 
5.05 



2.58 
2.95 
3.32 
3.68 
4.05 
4.42 
4.79 
516 
5.53 
5.89 



o.b i 
3 79 
4.21 
463 
5.05 
5.47 
5.89 
6.32 
6.74 



TABLE No. 1 6. Sizes of Numbers of the U. S. Standard 
Gage for Sheet and Plate Iron and Steel. 

(Brown & Sharpe Mfg. Co.) 





Approximate 


Approximate 


Weight Per 


Weight Per 


Number 


Thickness in 


Thickness in 


Square Foot 


Square Foot 


of 


Fractions of 


Decimal Parts 


in Ounces 


in Pounds 


Gage. 


an Inch. 


of an Inch. 


Avoirdupois. 


Avoirdupois. 


0000000 


% 


.5 


320 


20.00 


000000 


15 

3 2 


.46875 


300 


18.75 


00000 


7 
1 6 


.4375 


280 


17.50 


0000 


13 

3 2 


.40625 


260 


16.25 


000 


H 


.375 


240 


15. 


00 


1 1 

3 2 


.34375 


220 


13.75 





5 
1 6 


.3125 


200 


12.50 


1 


9 

3 2 


.28125 


180 


11.25 


2 


1 7 
6T 


.265625 


170 


10.625 


3 


X 


.25 


160 


10. 


4 


15 

64~ 


.234375 


150 


9.375 


5 


7 
3 2 


.21875 


140 


8.75 


6 


1 3 
64 


.203124 


130 


8.125 


7 


3 
1 6 


.1875 


120 


' 7.5 


8 


1 1 

6T 


.171875 


110 


6.875 


9 


5 

3 2 


.15625 


100 


6.25 


10 


9 
6T 


.140625 


90 


5.625 


11 


y? 


.125 


80 


5. 


12 


6¥ 


.109375 


70 


4.375 


13 


3 

3 2 


.09375 


60 


3.75 


14 


5 

6T 


.078125 


50 


3.125 


15 


9 

12 8 


.0703125 


45 


2.8125 


16 


1 
T6 


.0625 


40 


2.5 


17 


9 
16 


.05625 


36 


2.25 


18 


1 

20 


.05 


32 


2. 


19 


7 
160 


.04375 


28 


1.75 


20 


3 

8 


.0375 


24 


1.50 


21 


1 1 

3 20 


.034375 


22 


1.375 


22 


1 
3 2 


.03125 


20 


1.25 


23 


9 
3 20 


.028125 


18 


1.125 


24 


1 

¥o 


.025 


16 


1. 


25 


7 
3 20 


.021875 


14 


.875 


26 


3 

16 


.01875 


12 


.75 


27 


1 1 

64~0 


.0171875 


11 


.6875 


28 


1 
64" 


.015625 


10 


.625 


29 


9 
6¥0 


.0140625 


9 


.5625 


30 


1 
8 


.0125 


8 


.5 


31 


7 
640 


.0109375 


7 


.4375 


32 


1 3 
12 8 


.01015625 


Q/ 2 


.40625 


33 


3 

3 2 


.009375 


6 


.375 


34 


1 1 
12 8 


.00859375 


5^ 


.34375 


35 


5 
6¥0 


.0078125 


5 


.3125 


36 


9 
1210 


.00703125 


4^ 


.28125 


37 


1 7 
2 5 6 


.006640625 


4X 


.265625 


38 


1 

160 


.00625 


4 


.25 



(145) 



146 



WEIGHTS AND MEASURES. 



TABLE No. i7. Different Standards for Wire Gage in 
Use in the United States. 

Dimensions of Sizes in Decimal Parts of an Inch. 
(Brown & Sharpe Mfg. Co.) 



u- 6 

to 

SO 

- E 


American 

or Brown & 

Sharpe. 


Birmingham 

or 
Stubs' Wire. 


#6 * 


Trenton Iron 

Co., Trenton, 

N.J. 


«j 

u 

in 


U. S. Stand, 
for Plate. 


Number of 
Wire Gage. 


000000 












•46875 


000000 


00000 








.45 




•4375 


00000 


0000 


.46 


.454 


.3938 


.4 




•40625 


0000 


000 


.40964 


.425 


.3625 


.36 




•375 


000 


00 


.3648 


.38 


.3310 


.33 




•34375 


00 





.32486 


.34 


.3065 


.305 




•3125 





1 


.2893 


.3 


.2830 


.285 


.227 


•28125 


1 


2 


.25763 


.2S4 


.2625 


.265 


.219 


•265625 


2 


3 


.22942 


.259 


.2437 


.245 


.212 


•25 


3 


4 


.20431 


.238 


.2253 


.225 


.207 


•234375 


4 


5 


.18194 


.22 


.2070 


.205 


.204 


•21875 


5 


6 


.16202 


.203 


.1920 


.19 


.201 


.203125 


6 


7 


.14428 


.18 


.1770 


.175 


.199 


.1875 


7 


8 


.12849 


.165 


.1620 


.16 


.197 


.171875 


8 


9 


.11443 


.148 


.1483 


.145 


.194 


.15625 


9 


10 


1 .10189 


.134 


.1350 


.13 


.191 


.140625 


10 


11 


l .090742 


.12 


.1205 


.1175 


.188 


.125 


11 


12 


' .080808 


.109 


.1055 


.105 


.185 


.109375 


12 


13 


.071961 


.095 


.0915 


.0925 


.182 


.09375 


13 


14 


.064084 


.083 


.0800 


.08 


.180 


.078125 


14 


15 


•05706S 


.072 


.0720 


.07 


.178 


.0703125 


15 


10 


.05082 


.065 


.0625 


.061 


.175 


.0625 


16 


IT 


.045257 


.058 


.0540 


.0525 


.172 


.05625 


17 


18 


•040303 


.049 


.0475 


.045 


.168 


.05 


18 


19 


■03589 


.042 


.0410 


.04 


.164 


.04375 


19 


20 


.031961 


.035 


.0348 


.035 


.161 


.0375 


20 


21 


.028462 


.032 


.03175 


.031 


.157 


.034375 


21 


22 


.025347 


.028 


.028(5 


.028 


. 1 55 


.03125 


22 


23 


.022571 


•025 


.0258 


.025 


.153 


.028125 


23 


24 


.0201 


.022 


.0230 


.0225 


.151 


.025 


24 


25 


■0179 


.02 


.0204 


.02 


.148 


.021875 


25 


20 


.01594 


.018 


.0181 


.018 


.146 


.01875 


26 


27 


.014195 


.016 


.0173 


.017 


.143 


.0171875 


27 


2S 


.012641 


.014 


.0162 


.016 


.139 


.015625 


2S 


29 


.011257 


.013 


.0150 


.015 


.134 


.0140625 


29 


30 


.010025 


.012 


.0140 


.014 


.127 


.0125 


30 



WEIGHTS AND MEASURES. 



147 



TABLE No. 17. — (Continued). 



.imber of 
ire Gage. 


merican 
Brown & 
5harpe. 


mingham 

or 
bs' Wire. 


shburn & 
n Mfg. Co. 
orcester, 
Mass. 


nton Iron 
, Trenton, 
N.J. 


6 

IE 


C 1) 
TO « 

C/3 £ 


umber of 
ire Gage. 


*£ 





.3 3 




u 
HO 


uo 


£}■"" 


*£ 


31 


.008928 


.01 


0132 


.013 


.120 


.0109375 


31 


32 


.00795 


.009 


.0128 


.012 


.115 


.01015625 


32 


33 


.00708 


.008 


.0118 


.011 


.112 


.009375 


33 


34 


.006304 


.007 


.0104 


.01 


.110 


.00859375 


34 


35 


.005614 


.005 


.0095 


.0095 


.108 


.0078125 


35 


36 


.005 


.004 


.0090 


.009 


.106 


.00703125 


36 


37 


.004453 






.0085 


.103 


.006640625 


37 


38 


.003965 






.008 


.101 


.00625 


38 


39 


.003531 






.0075 


.099 




39 


40 


.003144 






.007 


.097 




40 



TABLE No. 18. Weight of Iron Wire in Pounds per 1 00 Feet. 



No. of 


American 


Birmingham 


No. of 


American 


Birmingham 


Wire 


or Brown & 


or 


Wire 


or Brown & 


or 


Gage. 


Sharpe. 


Stubs' Wire. 


Gage. 


Sharpe. 


Stubs' Wire. 


0000 


56.074 


54.620 


19 


0.341 


0.467 


000 


44.4683 


47.865 


20 


0.270 


0.324 


00 


35.265 


38.266 


21 


0.214 


0.271 


0. 


27.966 


30.634 


22 


0.170 


0.207 


1 


22.178 


23.850 


23 


0.135 


0.165 


2 


17.588 


21.373 


24 


0.107 


0.128 


3 


13.948 


17.776 


25 


0.0849 


0.106 


4 


11.061 


15.010 


26 


0.0673 


0.0858 


5 


8.772 


12.826 


27 


0.0534 


0.0678 


6 


6.956 


10.920 


28 


0.0423 


0.0519 


7 


5.516 


8.586 


29 


0.0335 


0.0447 


8 


4.375 


7.214 


30 


0.0266 


0.0381 


9 


3.469 


5.804 


31 


0.0211 


0.0265 


10 


2.751 


4.758 


32 


0.0167 


0.0214 


11 


2.182 


3.816 


33 


0.0132 


0.0169 


12 


1.730 


3.148 


34 


0.0105 


0.0129 


13 


1.372 


2.391 


35 


0.00836 


0.00662 


14 


1.088 


1.825 


36 


0.00662 


0.00424 


15 


0.863 


1.372 


37 


0.00525 




16 


0.684 


1.119 


38 


0.00416 




17 


0.542 


0.891 


39 


0.00330 




18 


0.430 


0.636 


40 


0.00262 





TABLE No. 10.— Decimal Equivalents of the Numbers of Twist Drill 
and Steel Wire Gage. (Brown & Sharpe Mfg. Co.) 



No. 


Size in 
Deci- 
mals. 


No. 


Size in 
Deci- 
mals. 


No. 


Size in 
Deci- 
mals. 


No. 


Size 

^ n . No. 
Deci- 
mals. 


Size 
in 
Deci- 
mals. 


No. 


Size in 
Deci- 
mals. 


1 


.2280 


15 


.1800 


29 


.1360 


42 


.0935 , 


55 


.0520 


68 


.0310 


2 


.2210 


16 


.1770 


30 


.1285 


43 


.0890; , 


36 


.0465 


69 


.02925 


3 


.2130 


17 


.1730 


31 


.1200 


44 


.0860; 


37 


.0430 


70 


.0280 


4 


.2090 


18 


.1695 


32 


.1160 


45 


.0820j 


38 


.0420 


71 


.0260 


5 


.2055 


19 


.1660 


33 


.1130 


46 


.08io; 


39 


.0410 


72 


.0250 


6 


.2040 


20 


.1610 


34 


.1110 


47 


.0785 


30 


.0400 


73 


.0240 


7 


.2010 


21 


.1590 


35 


.1100 


48 


.0760 


31 


.0390 


74 


.0225 


8 


.1990 


22 


.1570 


36 


.1065 


49 


.0730 


62 


.0380 


75 


.0210 


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In using the gages known as Stubs' Gages, there should be constantly borne in 
mind the difference between the Stubs Iron Wire Gage and the Stubs Steel Wire 
Gage. The Stubs Iron Wire Gage is the one commonly known as the English 
Standard Wire, or Birmingham Gage, and designates the Stubs soft wire sizes. The 
Stubs Steel Wire Gage is the one that is used in measuring drawn steel wire or drill 
rods of Stubs' make, and is also used by many makers of American drill rods. 

(148) 



<5eometr\>. 



Geometry is the science which teaches the properties of 
lines, angles, surfaces and solids. 

A point indicates only position and has neither length, 
breadth or thickness. A point has no magnitude. 

A line has length, but no breadth or thickness ; it is either 
straight, curved or mixed. 

A straight line is the shortest distance between two points. 

A curved line is continuously changing its position. 

A mixed line is composed of straight and curved lines. 

A surface has length and breadth, but no thickness ; it may 
be either plane or curved. 

A solid has length, breadth, and thickness or depth. 

An angle is the inclination of two lines which intersect or 
meet each other. The point of intersection is called the vertex 
of the angle. An angle is either right, acute or obtuse. 

A right angle contains 90 degrees. An acute angle contains 
less than 90 degrees. An obtuse angle contains more than 90 
degrees. 



Fig. 1. 




Fig. 3. 



Right Angle. Acute Angle. Obtuse Angle. 

* Polygons. 

Polygons are plane figures bounded on all sides by straight 
lines, and are either regular or irregular, according to whether 
their sides and angles are equal or unequal. The points at 
which the sides meet are called vertices of the polygon. The 
distance around any polygon is called the perimeter. 

A figure bounded by three straight lines, forming three 
angles, is called a triangle. 

The sum of the three angles in any triangle, independent of 
its size or shape, makes 180 degrees. 

All triangles consist of six parts ; namely, three sides and 
three angles. If three of these parts are known, one at least 
being a side, the other parts may be calculated. 

A triangle is called equilateral when all its three sides have 
equal length. Then all the three angles are equal, namely, 60 
degrees, because 60X3 = 180. (See Fig. 4). 

* Some authorities define as polygons only figures having more than four sides. 



i5o 



GEOMETRY. 



A triangle is called a right-angled triangle when one angle 
is 90 degrees; the other two angles will then together consist of 
90 degrees, because 90 + 90 = 180. ( See Fig. 5). 

An acute-angled triangle has all its angles acute. (See 
Fig. 6). 



Fig. 4. 



Fig. 6. 




Equilateral Triangle. Right-Angled Triangle. 



Acute Triangle. 



The longest side in a right-angled triangle is called the 
hypothenuse and the other two sides are called the base and 
perpendicular. The square of the length of the hypothenuse is 
equal to the sum of the squares of the lengths of the other two 
sides. ( See Fig. 5). 

a 2 + b 2 = c 2 

From this law the third side of a right-angled triangle can 
always be found, when the length of the other two sides is 
known. Thus : ( See Fig. 5). 



a = Vc 2 — b 2 



\/c 2 —a 2 c = Va 2 + b 2 



If, instead of the letters a, b, and c, numbers are used, for 
instance, a = 3 and b = 4 ; what then is the length of c ? 






a = \fh 2 — 4 2 
a = V L 2o — 10 
a = */ ( .)~ 



b = Vb 2 — 3 2 

b = \/25 — 9 

b = VJ6~ 
b = ± 



A square is a plane figure having four right angles and 
bounded by four straight lines of equal length. (See Fig. 7). 



Fig. 7. 



GEOMETRY. 



ISI 



A parallelogram is a plane figure whose opposite sides are 
parallel and of equal length. ( See Fig. 8). 

A rectangle is a parallelogram having all its angles right 
angles. ( See Fig. 9). 

A trapezoid is a plane figure bounded by four straight 
lines, of which only two are parallel. ( See Fig. 10). 



Fig, o. 



Fig. 9. 



Parallelogram. 



Fig. 10. 



Rectangle. 



Trapezoid. 



Fig. 11. 




A trapezium is a plane figure bounded by 
four sides, all of which have unequal length. 
(See Fig. 11). 

Polygons having four sides, and conse- 
quently four angles, are usually called quad- 
rangles. Polygons having more than four 
sides are named from the number of their 
sides. 



Trapezium. 



Thus : 






A polygon 


having 


five 


(( C( 


<< 


six 


u a 


a 


seven 


« u 


(C 


eight 


u u 


u 


nine 


u a 


a 


ten 


a u 


« 


eleven 


u u 


a 


twelve 



sides is called a pentagon. 

" a hexagon. 

" a heptagon. 

" an octagon. 

" a nonagon. 

" a decagon. 

" an undecagon. 

" a dodecagon. 

The sum of the degrees of all the angles of any polygon 
can always be found by subtracting 2 from the number of sides 
and multiplying the remainder by 180. 

For instance : 

The sum of degrees in any quadrangle is always (4 — 2) 
X 180 = 360 degrees. 

The sum of degrees in any pentagon will always be (5 — 2) 
X 180 = 540 degrees. 

This is a useful fact to remember in making drawings, as it 
may be used for verifying angles of polygons. 



*5 2 



GEOMETRY. 



Circles. 



Fig. 12 




Circle. 



The Circle is a plane figure bound- 
ed by a curved line called the circum- 
ference or perphery, which is at all 
points the same distance from a fixed 
point in the plane, and this point is 
called the center of the circle. ( See 
point c, Fig. 12). 

A Diameter is a straight line 
passing through the center of a circle or 
a sphere, terminating at the circumfer- 
ence or surface. (Sec line e-d, Fig. 12). 
A Radius is a straight line from the center to the circum- 
ference of circle or sphere. ( See line c-f, Fig. 12). 

Diameter = 2 X radius. The ratio of the circumference to 
the diameter of a circle is usually denoted by the Greek letter ?r 
and is expressed approximately by the number 8.1416 or 

2 2 Q 1 

- 7 Oj. 

Thus, if the circumference is required, multiply the diameter 
by 3.1416. If the diameter is required, divide the circumference 
by 3.1416. 

A Chord is a straight line terminating at the circumference 
of a circle but not passing through the center of the circle. (See 
line a-b, Fig. 12). The curved line a-b, or any other part of the 
circumference of a circle, is called an arc. 

Any surface bounded by the chord and an arc, like the 
shaded surface a-b, is called a segment. 

Any surface bounded by a chord and its two radii, like the 
shaded surface c-f-d, is called a sector. 



PROPERTIES OF THE CIRCLE. 

Circumference = Diameter X 3.1416 



Area 
Diameter 

Diameter 



= (Diameter) 2 X 0.7854 
= Circumference X 0.31831 



=V^ 



7854 

Diameter = 1.1283 X V area 

Circumference = 3.5449 X V area 

Length of any arc = Number of degrees X 0.017453 X radius. 
Length of arc of 1 Degree when radius is 1 is 0.017453. 
Length of an arc of 1 Minute when radius is 1 is 0.0002Q0888. 
Length of an arc of 1 Second when radius is 1 is 0.000004848. 
When the length of the arc is equal to the radius the angle 
is 57° 17' 45" = 57.2957795 degrees. 



TRIGONOMETRY. 



153 



TRIGONOMETRY. 

Trigonometry is that branch of geometry which treats of 
the solution of triangles by means of the trigonometrical 
functions. 

When the circumference of a circle is divided into 360 equal 
parts each part is called one degree. 

One fourth of a circle is 90 degrees = right angle, because 
4 X 90 = 360. ( See Fig. 13.) 

Fig. 13. fc h 





Fig. 14 

Circle = 4 right angles. 1 degree = 60 minutes (60'.) 

Circle = 360 degrees (360°.) 1 minute = 60 seconds (60".) 

Concerning the angle n ( see Fig. 14 ) the following are the 
trigonometrical functions : 

c g radius = 1. c h cosecant (cosec.) 

d b sine ( sin.) g f tangent ( tan.) 

c d cosine (cos.) k h cotangent (cot.) 
c f secant (sec.) 

The complement of an angle is what remains after sub- 
tracting the angle from 90°. Thus, the complement of an angle 
of 30° is 60° because 90 — 30 = 60. 

The supplement of an angle is what remains after subtract- 
ing the angle from 180°. Thus, the supplement of an angle of 
30° is 150° because 180 — 30 = 150. 

As all circles, regardless of their size, are divided into 360 
degrees, the trigonomical functions must always be alike if the 
radius and the angle that they denote are alike. 

It is on this basis that the tables of trigonometrical func- 
tions are calculated, and as radius is used the figure 1. 

In Table No. 20, the natural sine of 30° is given as 0.5 ; this 
means that if the line c g (see Fig. 14) is 1 foot, meter, or any 
other unit, and the angle n is 30 degrees, the line db will be 0.5 
of the same unit as the line eg. 

Sine 45° = 0.70711 ; that is, if the the angle n is 45 degrees 
and the line c g is 1 of any unit, the line d b is 0.70711 of the 
same unit. 

Cos. 30° = 0.86603 ; that is, if the angle n is 30 degrees and 
the line eg is 1 of any unit, the line a b or c d is 0,86603 of the 
same unit, 



J54 



TRIGONOMETRY. 



Sec. 30° = 1.1547 ; that is, if the angle n is 30 degrees and 
the line cgis 1 of any unit, the line cf is 1.1547 of the same unit. 
Cosec. 30° = 2 ; that is, if the angle ;/ is 30 degrees and the 
line eg is 1 of any unit, the line c h is 2 of the same unit. 

Tang. 30° = 0.57735 ; that is, if the angle n is 30 degrees 
and the line c g is 1 of any unit, the line g f is 0.57735 of the 
same unit. 

1.73205; that is, if the angle n is 30 degrees and 
1 of any unit, the line k h is 1.73205 of the 



30° = 
eg is 



Cot 
the line 
same unit. 

Increasing the angle n will increase sine, tangent and 
secant, but will decrease cosine, cotangent and cosecant. 

When the angle n approaches 90°, the tangents g f increase 
more and more to infinite length. When n actually reaches 90° 
of course c b coincides with c k and becomes parallel to gf, so 
that in an angle of 90° both the secant and the tangent have 
infinite length, which is denoted by the sign 00 ? and cosine and 
cotangent have vanished. 

In the first quadrant (that is when angle n does not exceed 
90°) the trigonometrical functions are all considered to be 
positive and are denoted by + (plus). When the angle n 
exceeds 90°, only sine and cosecants remain positive ; all the 
other functions have become negative and are denoted by — 
(minus). 

The following table gives the properties of the trigono- 
metrical functions in the four different quadrants : 



Degree. 


Sine. 


Cosine. 


0° to 90° 

90° to 180° 

180° to 270° 

270° to 360° 


Increase from to radius -\- 
Decrease from radius to 4- 
Increase from to radius — 
Decrease from radius to — 


Decrease from radius to -\- 
Increase from to radius — 
Decrease from radius to — 
Increase from to radius -f- 


Degree. 


Secant. 


Cosecant. 


0° to 90° 

90° to 180° 

180° to 270° 

270° to 360° 


Increase from radius to 0° -f- 
Decrease from oo to radius — 
Increase from radius to °° — 
Decrease from Oo to radius + 


Decrease from °° to radius -(- 
Increase from radius to °° 4- 
Decrease from oo to radius — 
Increase from radius to 0° — 



Degree. 


Tangent. 


Cotangent. 


0° to 90° 

90° to 180° 
180° to 270° 
270° to 360° 


Increase from to °° -j- 
Decrease from 0° to — 
Increase from to °° -\- 
Decrease from 0° to — 


Decrease from OO to -j- 
Increase from to °° — 
Decrease from °° to -f- 
Increase from0to°O — 



From the rule that the square of the hypothenuse is equal 
to the sum of the squares of the base and the perpendicular, it 
also follows that: 



TRIGONOMETRY. 



155 



Sin. 2 + cos. 2 = radius 2 . 
Tang. 2 -f- radius 2 = secant 2 . 
Cot. 2 -j~ radius 2 = cosecant 2 . 

But the trigonometrical tables are calculated with radius 
1, hence, 



sin. 



+ cos. 2 = 1. 



tang. 2 + 1 
cotang. 2 -j- 1 



= sec/ 

= cosecant 2 . 



tang. = 


sin. 


cosin. 


secant = 


1 


cosin. 


cotang. = 


cosin. 
sin. 




1 


cosec. — 


sin. 


cotang. = 


1 


tang. 



cosin. 



— \A — sin. 



Idllg 




cotang. 




secant 


tang, 
sin. 








1 




sin. 


cosec. 




. 




cosin. 




sin. 


cotang. 




sin. 


m. 


sin. 


COS. 2 


cos 


tang. 




Fig. 


16. 




Fig. 17. 




Sine and Cosine of the Sum of Two Angles. 

(See Fig. 15). 

Sine (a + b) = sin. a X cos. b -\- cos. a X sin. b. 
Cos. (a -j- b) = cos. a X cos. b — sin. a X sin. b. 

Sine and Cosine of Twice any Angle. 

(See Fig. 16). 

Sin. 2a = 2X sin. a X cos. a. 
Cos. 2 a = cos. 2 a — sin. 2 a. 

Sine and Cosine of the Difference of Two Angles. 

(See Fig. 17). 

Sin. (a — b) = sin. a X cos. b — cos. a X sin. b. 
Cosin. (a — b) = cos. a X cos. b + sin. a X sin. b. 



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TRIGONOMETRY. I 5 7 

The Trigonometrical Table and Its Use. 

Table No. 21 gives sine, cosine, tangent, and cotang, to 
angles from to 90 degrees with intervals of 10 minutes. 

For sine or tangent find the degree in the left-hand column 
and find the minutes on the top of the table. For instance, 
sine to 18° 40' = 0.3200G. 

If cosine or cotangent is wanted, find the degree in the 
column at the extreme right and the minutes at the bottom of 
the table. For instance, cotang 48° 10' = 0.89515. 

As the table only gives the angles and their trigonometrical 
functions with 10-minute intervals, any intermediate angle must 
be calculated by interpolations. For instance, find sine of 60° 
15' 10". 

Solution : 

Sine 60° 20' 0" = 0.86892 
Sine 60° 10' 0" = 0.86748 

Difference of 0° 10' 0" = 0.00144 
60° 15' 10" — 60° 10' 0" = 0° 5' 10' = 310 seconds and a 
difference of 10' = 600" increases this sine 0.00144. Therefore 
a difference of 310 seconds will increase the sine. 

310 X 0.00144 

m = 0.00074 

and sine 60° 10' 0" = 0.86748 

Therefore sine 60° 15' 5" = 0.86822 

Important. — During all interpolations concerning the 
trigonometrical functions, remember the fact that if the angle 
is increasing both sine and tangent are also increasing, and 
corrections found by interpolations must be added to the num- 
ber already found ; but as the cosine and cotangent decrease 
when the angle is increased, for these functions the corrections 
must be subtracted. 

Interpolations of this kind are not strictly correct, as neither 
the trigonometrical functions nor their logarithms differ in pro- 
portion to the angle. The error within such small limits as 
10 minutes is very slight. When very close calculations of 
great distances are required, tables are used which give the 
functions with less difference than 10 minutes; but for mechan- 
ical purposes in general these interpolations are correct for all 
ordinary requirements. It is very seldom in a draughting office or 
a machine shop that any angle is measured for a difference of 
less than 10'. 

To Find Secant and Cosecant of Any Angle. 

Divide 1 by cosine of the angle and the quotient is secant 
of the same angle. 

Divide 1 by sine of the angle and the quotient is cosecant 
of the same angle. 



158 



TRIGONOMETRICAL TABLES. 







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l66 TRIGONOMETRY. 

Logarithms Corresponding to the Trigonometrical 
Functions. 

Table No. 22 gives the logarithms corresponding to sine, 
cosine, tangent and cotang. for angles from to 90 degrees, 
with intervals of 10 minutes. For sine and tangent find the 
degree in the column to the left and the minutes at the top of 
the table. For instance : 

Log sine 19° 30' = 9.523495 — 10. 

This, of course, is also logarithm to the fraction 0.33381, 
which is sine of 19° 30'. 

For cosine and cotang. find the degree in the column to the 
extreme right in the table, and find the minutes at the bottom 
of table. For instance: 

Log. cotang. 37° 10' = 10.120259 — 10 = 0.120259. 

Note. — In this table the index of the logarithm is increased 
by 10, therefore — 10 must always be annexed in the logarithm. 

Logarithms to angles between those in the table may be 
obtained by interpolations. For instance, find log. sine 25° 45'. 

Solution : 

Log. sine 25° 50' = 9.639242 — 10 

Log. sine 25° 40' = 9.636623 — 10 

Difference 0.002619 

This difference in the logarithm corresponds to a difference 
in this angle of 10 minutes ; therefore a difference of 5 minutes 
in the angle will make a difference of 0.001309 in the logarithm. 
Thus : 

Log. sine 25° 40' = 9.636623 — 10 
Difference 5' = 0.001309 

Log. sine 25° 45' = 9.637932 — 10 

Example 2. 

Find angle corresponding to logarithmic sine 9.894246 — 10. 

Solution : 

In the table of logarithms of sine : 

9.894546 — 10 corresponds to 51° 40' 
9.893544 — 10 corresponds t o 51° 30' 

Difference 0.001002 corresponds to 0° 10' 

To logarithm 9.894246 — 10 must, therefore, correspond an 

angle somewhere between 51 c 30' and 51° 40', which is found 

thus : 

The given logarithm is 9.894246 — 10 

Nearest less logarithm 9.893544 — 10 for 51° 30' 

Difference 0.000702 

Therefore, the correction to be added to the angle already 
found will be : 

0.000702 x io _ 

0.001002 ~~ u ' 

Thus, the logarithmic sine 9.894246 — 10 gives 51° 37' 



TRIGONOMETRY. 1 67 



Example 3. 

Find log. to tangent of 50° 45' 
Solution : 

Log. tangent 50° 50' = 0.089049 
Log. tangent 50° 40' = 0.086471 



Difference 0° 10' = 0.002578 in the logarithm. There- 
fore a difference of 5' in the angle will give 0.001289 in the 
logarithm. 



Thus: 



Log. tangent 50° 40' = 0.086471 
Difference 5' = 0.001289 



Log. tangent 50° 45' = 0.087760 
Example 4. 

Find the angle corresponding to log. tangent 9.899049 — 10. 

Solution : 

Log. tangent 38° 30' = 9.900605 — 10 

Log. tangent 38° 20' = 9.898010 — 10 

Difference 0° 10' corresponds to 0.002595 

The given logarithm = 9.899049 — 10 

Nearest less logarithm = 9.898010 — 10 gives 38° 20' 

Difference = 0.001039 

The difference to be added to the angle already found will 
0,00 1039 x io _ 
ue 0.002595 ~~ u * * 

The tabulated logarithm 9.898010 — 10 gives angle 38° 20' 
Difference 0.001039 gives angle 4' 

Logarithm 9.899049 — 10 gives angle 38° 24' 

To Find Logarithm for Secants and Cosecants. 

Logarithm for secants is found by subtracting log. cosine 
from log. 1. 

For instance, find logarithmic secant 30°. 
Solution : 

Log. 1 = 10.000000 — 10 
Log. cosine 30° = 9.937531 — 10 

Log. secant 80° = 0.062469 

Logarithm for cosecants is found by subtracting log. sine 
from log. 1. For instance, find logarithmic cosecant 35°. 
Solution: 

Log. 1 = 10.000000 — 10 
Log. sine 35° = 9.758591 — 10 
Log. co-secant 35° = 0.241409 

Note. — What is said concerning interpolations of trigono- 
metrical functions in general in the note headed "Important" 
on page 157, will also apply to their logarithms. 



1 68 



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176 



TRIGONOMETRY. 



Solutions of Right=Angled Triangles. 

Fig. 18. 



Right-angled triangles (see Fig. 18) 
may be solved by the following for- 
mulas : 



Solving for Any Side. 



C 



cosec. a 

C 
sec. a 

A 
sine « 



B 

cot. a 

A 
tang, a 
JB_ 
cos. a 



A = C X sine a = B X tang, a = 
B = C X cos. a = A X cot. a = 

C = A X cosec. a = B X sec. a = 

A = VC 2 — B 2 
B= VC 2 —A 2 
C = VA 2 + B 2 

Solving for Any Function or for Any Angle. 



Sin. 


a 


A 
— C 


-r A 
Tang. a = ^~ 


Sec. a = 


c 

' B 


Cos. 


a 


B 

— C 


Cot. a = ~^- 


Cosec. a = 


C 

~aT 


Sin. 


a 


= cos. b 


Tang. a = cot. b 


Sec. a = 


cosec. b 


Sin, 
Sin. 


b 
■b 


= cos. a 

B 

— ~~c~ 


Tang, b = cot. a 
Tang, b — -j- 


Sec. b = 
Sec. b = 


cosec. a 
c 

A 


Cos. 


b 


A 
— C 


Cot. b = —g- 


Cosec. b = 


C 

B 






Angle a = 


d0° — b. " Angl 
Solving for Area. 


e b = 90° — 


a. 






Area = 


A X B 

2 










C 2 X sin. a X 


cos. a C 2 X cos. t 


j X sin. b 






2 


— 2 








B 2 X tang, a 


A 2 X tang. 


b 





TRIGONOMETRY. 



177 



Fife. 19 




Example. 

Find angles a and b and the 
side X in the right-angled triangle. 
(Fig. 19). 



** 12.5 feet * 



Tangent corresponding to a = -j^5~ = 0-4 



Tangent corresponding to b = 



12.5 



= 2.5 



By the trigonometrical table the angles are obtained thus : 
Tangent 0.40000 gives 21° 48' 
Tangent 2.50000 gives 68° 12' 

Therefore : 

Angle a = 21° 48' and angle b = 68° 12[. 
Angle b may also be found by subtracting angle a from 
90°, thus : 

Angle b = 90° — 21° 48' = 68° 12' 
The length of the side X may be found thus : 
5 



x 



X 



sin. c 
5 



thus 



0.37137 

x = 13.464 feet long. 
By means of logarithms the length of the side x is obtained 

Log. x = log. 5 — log. sin. 21° 48' 
Log. x — 0.698970 — (9.569804 — 10) 
Log.x— 1.129166 

x = 13.464 feet long. 

Solution of ObIique=Angled Triangles. 



Fig. 20. 




Oblique-angled triangles 
(see Figs. 20-21-22) may be solved 
by the following formulas : 



Fig. 21. 





I7& TRIGONOMETRY. 

Solving for Any Side. 

. Csin. a B sin. a l~ " 

A = -STT = -riO- - <yjB*+C*-2BCcos.a 

_, C sin. b A sin. b I ~ 

B = -^nT^^^i,^- = ^C 2 + A 2 -2ACcos.b 

-, A sin. c B sin. c \ 

C =-^^- = -^7- =^A* + £*-2ABcos.c 
Solving for Any Angle. 



Cos. a = 



_ B 2 + C 2 — A 2 



2BC 



CQ ^ b = A 2 +C 2 -B 2 
2 A C 

Cos.c= A2 + B2 - C 2 
2 A B 

Sin. a = sin. <5 = sin. c — 

B C 

B /? 

Sin. b = sin. c = sin. a — — 

C A 

C C 

Sin. ^ = sin. a = sin. b — 

A B 

a = 180° — (b-+ c) 
b = 180° — (a±c) 
c =180° — (a+ b) 

Solving for Area. 

Area _ sin, c X A X B _ sin, a X C X B __ sin, b X A X C 

2 2 2 

Example 1. 

Find the length of the side C (see Fig. 20) when angle a = 
20° 38' 12", angle c — 117° 48' 5", and side A — 12.75 feet long. 

Note. — The angle c exceeds 90°, therefore the supplement 
of the angle must be used, which is 180° — 117° 48' 5" = 
62° 11' 55". 

Thus the solution : 

12.75 X sin. 62° 11' 55" 



C — 
C= 



sin. 20° 38' 12" 
12.75 X 0.88456 



0.35243 
C= 32 feet long. 



trigonometry. i7§ 

Example 2. 

Find the length of the side B (see Fig. 20) when angle b 
is 41° 33' 43", side C is 32 feet and side A is 12.75 feet. 

In this example two sides and their included angle are 
given and the third side is required ; therefore the formula 

B = *JA 2 + C 2 — 2 A C cos. b must be used. 
Solution : 

B = Vl2.75 2 + 32 2 — 2 X 12.75 X 32 X 0.748238 

B — Vll86.562 — 610.562 

B — V576 == 24 feet long. 

Example 3. 

Find the length of the side B when side A is 12.75 feet 
long, angle b is 41° 33' 43" and angle c is 117° 48' 5' . ( See 
Fig. 20). _ 

In this problem one side and its two adjacent angles are 
given ; therefore it can not be solved directly by any of the 
preceding formulas, but the first thing to do is to find the angle 
opposite to side A. 

Thus: Angle a = 180° — (41° 33' 43" -f 117° 48' 5") = 
20° 38' 12". The side B may be found by the formula 

A sin. b 

B = • „ 



Solution : 
B = 

B = 



A sin. 41° 33' 43" 
sin. 20° 38' 12" 
12.75 X 0.66343 



0.35242 

B = 24 feet long. 
Example 4. 

Find length of the side C when B is 24 feet long, angle c is 
117° 48' 5" and the side A is 12.75 feet long. ( See Fig. 20). 
Solution : 

C = VA 2 + B 2 — 2AB cos. c 

C = Vl2.75 2 — 24 2 — 2 X 12.75 X 24 X (—0.4664) 

C = Vl62.56 + 576 + 285.44 

C = Vl02T = 32 feet long. 

Note. — In this example the cos. of 117° 48' 5" is used, 
which, in numerical value, is equal to cos. of 62° 11' 55" = 
0.4664, but cos. in the second quadrant is negative (see page 154); 
therefore cos. 117° 48'' 5" = ( — 0.46645) and the essential sign 
of the last product after it is multiplied by this negative cos. 
must change from — to +. ( See Algebra, page 63). 



I So TRIGONOMETRY. 

Example 5. 

Find the length of the side A when C is 32 feet long, angle 
a is 20° 38' 12" and angle c is 117° 48' 15". 
Note. — Supplement to c is 62° 11' 55". 

Solution : 

C sin. a 



A = 
A = 



sin. c 
32 X 0.35242 



0.88456 

A = 12.75 feet long. 

In this example, as in the preceding one, we use the sup- 
plement of the angle in obtaining its function, but here it has no 
influence on the signs because sin. is positive as well in the 
second as in the first quadrant. 

Example 6. 

Find angle a in Fig. 20, when A is 12.75 feet, B is 24 feet 
and C is 32 feet. 

Solution : 

_ B 2 + C 2 — A 2 



cos. a 



2BC 



24 2 + 32 2 — 12.75 2 
cos. a = 



cos. a 



2 X 24 X 32 

_ 576 + 1024 — 162.5625 
1536 

cos. a — 0.93583 
Angle a = 20° 38' 12" 

Example 7. 

Find angle b, Fig. 20, by the same formula. 

Solution : 

A i + C 2 — B 2 



cos. b = 



cos. b = 



cos. b = 



2AC 
12.75 2 4- 32 2 — 24 2 

2 X 12.75 X 32 
610.5625 



SI 6 

cos. & = 0.748238 
Angle b = 41° 33' 43" 



TRIGONOMETRY. l8l 



Example 8. 

Find angle c, Fig. 20, by the same formula. 
Solution : 

_ A 2 + B 2 — C 2 



COS. c 



cos. c = 



COS. c = 



2AB 

12.75 2 + 24 2 — 32 2 
2 X 12.75 X 4 

738.5625 — 1024 



612 

cos. c = — 0.46640 

Supplement to angle c = 62° 11' 55", and angle c = 117° 48' 5" 

Note. — The negative cosine indicates that it is in the sec- 
ond quadrant, therefore the angle is over 90°. 

The angle corresponding to this cosine is the supplement 
of angle c. To obtain angle c, the angle of its supplement must 
be subtracted from 180°. 

Example 9. 

Find angles a, b and c in Fig. 20, when side A is 12.75 
feet, B 32 feet, and C 24 feet. 

B% + C 2 — A 2 
cos. a — - ^ c 

2BC 
cos.tf = 32 '+ 242 - 12 - 752 



cos. a 



2 X 32 X 24 
1437.4375 



1536 

cos. a = 0.93583 
Angle a = 20° 38' 12" 

Angle b may be found by the formula : 

r> 

sin. b = sin. a J— 
A 

sin. b — sin. 20° 38' 12" JL 

A 

sin. b = 0.35244 X 



12.75 

sin. b — 0.35244 X 1.8824 
sin. b — 0.66343 
Angle b = 41° 33' 43" 



I 8 2 TRIGONOMETRY. 

Angle c may be found by the formula: 

c = 180° — (a + b) 

c = 180° — (20° 38' 12" + 41° 33' 43") 

c = 180° — 62° 11' 55" 

c — 117° 48' 5" 
Example 10. 

Find the area of a triangle (see Fig. 20), when it is known 
that side A is 12.75 feet, side B is 24 feet, and the including 
angle C— 117° 48' 5". 

Solution : 

Sin. to supplement of 117° 48' 5" = sin. 62° 11' 55" = 
0.88456. 

Area = sin " C X A X B 

2 

A 0.88456 X 12.75 X 24 iOK OA c . 

Area = — — = 135.34 square feet. 

Example 11. 

Find angle c and the sides X and y in the triangle, Fig. 23. 

Solution : 

C — 180° — (40° + 60°) = S0 C 

ThesideX= 25Xsin - 40 ° 



X — 
X = 



sin. 60° 
25 X 0.64279 

0.86603 
16.06975 




0.86603 ^ y 

X = 18.556 meters long. 
By the use of logarithms the side Xis solved thus: 
Log. X = log. 25 -f log. sin. 40° — log. sin. 60°. 
Log. X— 1.39794 + (9.808067—10) — (9.937531—10). 
Log.X^ 1.268476 

X = 18.556 meters long. 

r-* -1 25 X sin. 80° 

The side y = 



y 



sin. 60° 
25 X 0.98481 



0.86603 
y = 28.429 meters long. 
By the use of logarithms the side y is solved thus: 

Log. y = log. 25 + log. sin. 80° — log. sin. 60°. 
Log. y — 1.39794 4- (9.993351—10) — (9.937531—10). 
Log. y = 1.45376 

y = 2S.429 meters long. 




•* 42 meter >J 



TRIGONOMETRY. 1 83 

Example 12. 

Find angles c and b and the length 
of the side X in Fig. 24. 

Sin. c = 42Xsin - 54 ° 
35 

Sin. c = 42X0.80002 
35 

Sin. c — 0.97082 
Angle c = 76° 7' 26" 
Angle b = 180° — (54° 0' 0" + 76° V 26") = 49° 52' 34" 

Side X — 35 X sin. 49° 52' 34" 
sin. 54° 

X= 35 X °- 76465 - 33.08 meters long. 
0.80901 

By means of logarithms the side X is solved thus : 

Log. X — log. 35 + log. sin. 49° 52' 34" — log. sin. 54°. 
Log. X= 1.544068 + (9.883463—10) — (9.907958—10). 
Log.X= 1.519573 

X = 33.08 meters long. 

Note. — The angle c is obtained by interpolation thus : In 
the table of trigonometrical functions the sine 0.97100 corre- 
sponds to the angle 76° 10' and the sine 0.97030 corresponds to 
the angle 76°. Thus, a difference of 0.00070 in the sine gives 
a difference of 10' = 600" in the angle. 

The sine to angle c is 0.97082 
The nearest less sine in the table is 0.97030 corresponding 
to angle 76° 0' 0". Difference, 0.00052 

Therefore when an increase in sine of 0.00070 corresponds 
to an increase of 600" in the angle, an increase of 0.00052 will 

increase the angle 60Q X - 00052 ^ 446" = 0° 7' 26" 

0.00070 

thus, the angle corresponding to the sine 0.97082 must be 
76° 7' 26". 



1 84 



GEOMETRY. 



% 



^ 



ye 



>f 



y s 



TV 



PROBLEMS IN GEOMETRICAL DRAWING. 



To divide a straight line into a 
given number of equal parts. (See 
Fig. 1). 

Given line a b, which is to be 
divided into a given number of equal 
parts. Draw the line b c, of indefinite 
length, and point off from b the re- 
quired number of equal parts, as h,g, 



Fig. 1 



f, e, d, c'\ join c' and a, and draw the other lines parallel to c' a. 



XV 



Fig. 



To erect a perpendicular at a 
given point on a straight line. 
(See Fig. 2). 

Given line a b, and the point 
x. The required perpendicular 
is xy. 

Solution : 

With x as center and any 
With 1 and 2 as 



■l A '2 

radius, as xl, cut the line a b at 1 and 2. 
centers and with a radius somewhat greater than 1 to x, describe 
arcs intersecting each other at y. Draw x y. This will be the 
required perpendicular. 

From a given point without a straight line to draw a per- 
pendicular to the line. (See Fig. 3). 

Given line a b and the point c. 
The required perpendicular is x. 

Solution: 

With the point c as center and any 
radius as c 1, strike the arc 1 to 2. With 
1 and 2 as centers and any suitable 
radius, describe arcs intersecting each 
other at «, lay the straight edge through 
points n and c and draw the perpendicular x. 

To erect a perpendicular at the extremity of a straight line. 
(See Fig. 4). 

The required perpendicular is x. 

Solution : 

From any point, as c, with radius 
as a c, draw the circle. From point 
of intersection, ;/, through center, c, 
draw the diameter ?i p. From the 
points, through the point of intersec- 
tion at ft, draw the perpendicular x. 

The correctness of this con- 
struction is founded on the principle 
that inside a half circle no other 




Given line a b. 
Fig. a 



nC 



V 



\ 



yn. 



PROBLEMS IN GEOMETRICAL DRAWING. 



'85 



angle but an angle of 90° can simultaneously touch three points 
in the circumference when two of these points are in the point 
of intersection with the diameter and 
the circumference and the third one 
anywhere on the circumference of the 
half circle. The pattern maker is mak- 
ing practical use of this geometrical 
principle, when he by a common car- 
penter's square is trying the correctness 
of a semi-circular core box, as shown 
in Fig. 5. 




Draw a line parallel to a 
given line. (See Fig. 6). 

Given line a b. The required 
line x y. 

Solution : 

Describe with the compass « — 
from the line a b, the arcs 1 and 
2 ; draw line x y, touching these arcs. 



x -- „ ^. 



Fig. 6 



To divide a given angle into two 
equal angles. 

The given angle, a b c, is divid- 
ed by the line b d. 

Solution : 

With b as center and any radius, 
as b 1, describe the arc 1 to 2. With 
1 and 2 as centers and any suitable 
radius, describe arcs cutting each 
other at d. Draw line b d, which 
will divide the angle into two equal parts. 



Fig. 7 




To draw an angle equal to a given angle 

Given angle a b c. Construct 
angle x y z. 

With b as center and any radius, 
as b 1, describe the arc 1 to 2. 
Using y as center and without alter- 
ing the compass, describe the arc /, 
intersecting y z. Measuring the 
distance from 2 to 1 on the given 
angle, transfer this measure to the 
arc /, through the point of intersection. Draw the line y x, 
and this angle will be equal to the first angle. 

Note. — Angles are usually measured by a tool called a pro- 
tractor, looking somewhat like Fig. 9 or 10, usually made from 
metal, and supplied by dealers in draughting instruments. A 




i86 



PROBLEMS IN GEOMETRICAL DRAWING. 



protractor may also be constructed on paper and used for 
measuring angles, but it should then always be made on as 
large a scale as convenient. 



Fig 





To draw a protractor with a division of 5°. (See Fig. 10). 

Construct an angle of exactly 90 degrees, divide the arc into 
nine equal parts, then each part is 10 c ; divide each part into 
two equal parts and each is 5°. 



Prove that the sum of the three angles in a triangle consists 

of 180°. (See Fig. 11). 

a F,G - 11 Solution: 

In the triangle a b c, extend 
the base line to i. Draw the line 
o ft, parallel to the side a b, 
thereby the angles will be equal 
to the angle d, and the angle h 
must be equal to angle c. The 
angle/" is one angle in the triangle 
and /+ g + h = 180°, therefore 
/ + d + c must also be 180°. 




To draw on a given base line a 
30° and 60°. (See Fig. 12). 



triangle having angles 90 c 



Given line a b, required triangle is a, c, b. 

Solution: 



Fig. 



12 




Extend the line a b to twice its 
length, to the point e. With e and b as 
centers strike arcs intersecting each 
other and erect the perpendicular a c. 
With b as center and any radius as /, 
draw the arc / m. With / as center 
and with the same radius, describe arc 
intersecting at m. From b through 
point of intersection at m, draw line b 
intersecting the perpendicular at c. 
This will complete the triangle. 



PROBLEMS IN GEOMETRICAL DRAWING. 



I8 7 



To draw a square inside a 
given circle. (See Fig. 13). 

Solution : 

Draw the line a b through the 
center of the circle. From points 
of intersection at a and 6, describe 
with any suitable radius arcs inter- 
secting at/j and ///. Draw through 
the points the line c d. Connect 
the points of intersection on the 
circle and the required square is 
constructed. 




To draw a square outside a 
given circle. (See Fig. 14). 

Solution : 

Draw lines a b and c d, and 
from points of intersection at b and 
c, describe half circles ; their points 
of intersection determine the sides 
of the square. 



To draw a hexagon within a given 
circle. (See Fig. 15). 

Apply the radius as a chord succes- 
sively about the circle ; the resulting 
figure will be a hexagon. 




Fig. 15. 



To inscribe in a circle a regular polygon of _ 

any given number of sides. 

Solution : 

Divide 360 by the number of sides, and the 
quotient is the number of degrees, minutes, and 
seconds contained in the center angle of a triangle, 
of which one side will make one of the sides in 
the polygon. For instance, draw a hexagon by this method. 
(See Fig. 16). 360 = 6QO 

6 




iSS 



PROBLEMS IX GEOMETRICAL DRAWING. 



Fig. 17 




To find the center in a given 
circle. (See Fig. IT). 

Solution: 

Draw anywhere on the circumfer- 
ence of the circle two chords at ap- 
proximately right angles to each other, 
bisect these by the perpendiculars x 
and y, and their point of intersection 
is the center of the circle. 



To draw any number of 
circles between two inclined 
lines touching themselves and 
the lines. (See Fig. IS). 

Solution: 

Draw center line ef. Draw 
first circle on line i g. From 
point of intersection between 
this circle and the center line draw the line //. perpendicular to 
a b. Describe with a radius equal to //, the arc intersecting at 
g 1 . draw line^ 1 i 1 , parallel to g i. and its point of intersection 
with the center line gives the center for the next circle, etc. 




Fig 



To draw a circle through 
three given points. (See Fig. 19). 
The given points are a. b, and c. 

Solution: 

From a and b as centers with 
suitable radius, describe arcs inter- 
secting at e e. Draw aline through 
these points. From b and c as cen- 
ters, describe arcs intersecting at d 
d\ draw a line through these points. 

The point where these two lines intersect is the center of the 

circle. 





To draw two tangents to a 
circle from a given point without 
same circle. (See Fig. 20). 

Given point a, and the circle 
with the center n. The required 
tangents are a d, and a b. 

Solution : 

Bisect line n a. With c as 
center and radius a c, describe 



PROBLEMS IN GEOMETRICAL DRAWING. 



89 



the arc b d through the center of the circle. The points of 
intersection at b and d are the points where the required tan- 
gents a b and a d will touch the circle. 




Fig. 22. 




To draw a tangent to a given point in 
a given circle. (See Fig. 21). 

Given circle and the point k, x y is 
required. 

Solution : 

The radius is drawn to the point h and 
a line constructed perpendicular to it at the 
point h. This perpendicular, touching the 
circle at k, is called a tangent. 

To draw a circle of a certain size that 
will touch the perphery of two given cir- 
cles. (See Fig. 22). 

Given the diameter of circles a, b, 
and c. Locate the center for circle c, 
when centers for a and b are given. 

Solution : 

From center of #, describe an arc 
with a radius equal to the sum of radii of a and c. From b as 
center, describe another arc using a radius equal to the sum of 
the radii of b and c. The point of intersection of those two 
arcs is the center of the circle c. 

Note. — This construction is useful when locating the center 
for an intermediate gear. For instance, if a and b are the pitch 
circles of two gears, c would be the pitch circle located in correct 
position to connect a and b. 

To draw an ellipse, the longest 
and shortest diameter being given. 
The diameters a b and c d are 
given. The required ellipse is 
constructed thus : (See Fig. 23). 

From c as center with a radius 
a n, describe an arc f x f. The 
points where this arc intersects a 
b are foci. The distance fn is 
divided into any number of parts, 
as 1, 2, 3, 4, 5. With radius 1 to b, 
and the focus/ as center, describe arcs 6 and 6 1 ; with the same 
radius and with f 1 as center describe arcs 6 2 and 6 3 . With 
radius 1 to a and/" 1 as center, describe arcs intersecting at 6 and 
6 1 ; with the same radius and with f as center, describe arcs 
intersecting at 6 2 and 6 3 . Continue this operation for points 2, 
3, etc., and when all the points for the circumference are in this 




>—-\b 



190 



PROBLEMS IN GEOMETRICAL DRAWING. 



way marked out, draw the ellipse by using a scroll. It is a 
property with ellipses that the sum of any two lines drawn from 
the foci to any point in the circumference is equal to the largest 
diameter. For instance : 

/1 e +fe, = ab, or/6 1 +/* 6 1 , = a b. 




Cycloids. 

Suppose that a round disc, c, rolls on a straight line, a b, and 
that a lead pencil is fastened at the point r ; it will then describe 

a curved line, a, /, r, ;/, b. This 
line is called a cvcloid. (See 
Fig. 24). 

This supposed disk is usual- 
ly called the generating circle. 
The line a b is the base line of 
the cycloid and is equal in length 
to 7T times m r, or practically 3.1416 times the diameter of the 
generating circle. The length of the curved line a, /, r, n, b, is 
four times r m, (four times as long as the diameter of the 
generating circle). 

A circle rolling on a straight line generates a cycloid. 
(See Figs. 24 and 25). 

A circle rolling upon another circle is generating an 
epicycloid. (See Fig. 26). 

A circle rolling within another circle generates a hypo- 
cycloid. (See Fig. 27). 

To draw a cycloid, the generating circle being given. 

Solution : 

Divide the diameter of the 
rolling circle in 7 equal parts. 
Set oft 11 of these parts on each 
side of a on the line d e. This 
will give a base line practically 
equal to the circumference. 
Divide the base line from the 
point a into any number of equal parts; erect the perpendicu- 
lars, with center-line as centers and a radius equal to the radius 
of the generating circle describe the arcs. On the first arc from 
d or e set off one part of the base line. On the second arc set 
off two parts of the baseline ; on the third arc three parts, etc, 
This will give the points, through which tQ draw the cycloid, 




PROBLEMS IN GEOMETRICAL DRAWING. 



I 9 I 



To draw an epicycloid (see Fig. 26), the generating circle a 
and the fundamental circle B being given. 



Solution 



Fl.G. 26 




Concentric with the circle B, describe 
an arc through the center of the generating 
circle. Divide the circumference of the 
generating circle into any number of equal 
parts and set this off on the circumference 
of the circle B. Through those points draw 
radial lines extending until they intersect 
the arc passing through the center of the 
generating circle. These points of inter- 
section give the centers for the different positions of the gener- 
ating circle, and for the rest, the construction is essentially the 
same as the cycloids. In Fig. 26, the generating circle is shown 
in seven different positions, and the point n, in the circumfer- 
ence of the generating circle, may be followed from the position 
at the extreme left for one full rotation, to the position where it 
again touches the circle B. 



To draw a hypocycloid. (See Fig. 27). 

The hypocycloid is the line generated 
by a point in a circle rolling within another 
larger circle, and is constructed thus: (See 
Fig. 27). 



Fig. 27. 




Divide the circumference of the gener- 
ating circle into any number of equal parts. 
Set off these On the circumference of the 
fundamental circle. From each point of 
division draw radial lines, 1, 2, 3, 4, 5, 6. 
From 11 as center describe an arc through 
the center of the generating circle, as the 

arc c d. The point of intersection between this arc and the 
radial lines are centers for the different positions of the gener- 
ating circle. The distance from 1 to a on the fundamental 
circle is set off from 1 on the generating circle in its first new 
position ; the distance 2 to a on the fundamental circle is set off 
from 2 on the generating circle in its second position, etc. For 
the rest, the construction is substantially the same as Figs. 25 
and 26. 



Note. — If the diameter of the generating circle is equal to 
the radius of the fundamental circle, the hypocycloids will be a 
straight line, which is the diameter of the fundamental circle. 



192 



PROBLEMS IN GEOMETRICAL DRAWING. 




Involute. 

An involute is a curved line which may be assumed to be 
generated in the following manner : Suppose a string be placed 

around a cylinder from a to b, in the 

f ig. 2 8 _ p A direction of the arrow (see Fig. 28), 

and having a pencil attached at b\ 
keep the string tight and move the 
pencil toward c, and the involute, 
b c, is generated. 

To draw an involute. 
Solution : 

From the point b, (see Fig. 28) 
set off any number of radial lines at 
equal distances, as 1, 2, 3, 4, 5. From 
points of intersection draw the tangents (perpendicular to the 
radial lines). Set off on the first tangent the length of the arc 1 
to b ; on the second tangent the arc 2 to b, etc. This will give 
the points through which to draw the involute. 

To draw a spiral from a given 
point, c. 

Solution : 

Draw the line a b through the 
point c. Set off the centers r and 
S, one-fourth as far from c as the 
distance is to be between two lines 
in the spiral. Using r as center, 
describe the arc from c to 1, and 
using S as center, describe the arc 
from 1 to 2 ; using r as center, de- 
scribe the arc from 2 to 3, etc. 



Fig. 29 




If a cone (see 




Conical Sections. 

Fig. 30), is cut by a plane on the line a b, 
which is parallel to the center line, the 
section will be a hyperbola. 

If cut by a plane on the line c d, which 
is parallel to the side, the section will be a 
Parabola. 

If cut by a plane on the line^, which 
is parallel to the base line, the section will 
be a circle. 

If cut by a line, e f, which is 
neither parallel to the side, the center- 
line nor the base, the section will be an 
ellipse. 



MENSURATION. 



193 




-2 :feet — 



MENSURATION. 

If each side in a square ( see Fig. 1 ) is two feet long, the 
area of the figure will be 4 square feet; that is, it contains four 
squares, each of which is one square foot. 
Thus the area of any square or rectangle is fig. 1 • 

calculated by multiplying the length by the 
width. 

Example 1. 

What is the area of a piece of land 
having right angles and measuring 108 feet 
long and 20 feet wide ? 

Solution : 

108 X 20 = 2160 square feet. 

Example 2. 

What is the area in square meters of a square house-lot 80 
meters long and 30 meters wide. 

Solution : 

30 X 30 == 900 square meters. 

( Square meter is frequently written //z 2 and cubic meter is 
written m s ). 

A square inscribed in a circle is half in area of a square 
outside the same circle. Divide the side of a square by 0.8862, 
and the quotient is the diameter of a circle of the same area as 
the square. 



The Difference between One Square Foot and One Foot 

Square. 

One foot square means one foot long and one foot wide, 
but one square foot may be any shape, providing the area is one 
square foot. For instance, Fig. 1 is two feet square, but it con- 
tains four square feet. One inch square means one inch long 
and one inch wide, but one square inch may be any shape, 
provided the area is one square inch. One mile square means 
one mile long and one mile wide, but one square mile may have 
any shape, provided the area is one square mile. 



Area of Triangles. 

The area of any triangle may be found by multiplying the 
base by the perpendicular height and dividing the product by 2. 



194 mensuration. 

Example. 

Find the area of a triangle 16 inches long and 5 inches per- 
pendicular height 
Solution ; 

Area = = 40 square inches. 

The perpendicular height in any triangle is equal to the 
area multiplied by 2 and the product 'divided by the base. 

The area of any triangle is equal to half the base multiplied 
by the perpendicular height 

The perpendicular height of any equilateral triangle is 
equal to one of its sides multiplied by 

The area of any equilateral triangle may be found by mul- 
tiplying the square of one of the sides by 0.433. 

Example. 

Find the area of an equilateral triangle when the sides are 
12 inches long. 

Solution : 

-Area = 12 X 12 X 0.433 = '32.352 

The side of any equilateral triangle multiplied by 0.6582 
gives the side of a square of the same area. 

The side of any equilateral triangle divided by 1.3468 gives 
the diameter of a circle of the same area. 



To Figure the Area of Any Triangle when Only the 
Length of the Three Sides is Given. 

Rule. 

From half the sum of the three sides subtract each side 
separately: multiply these three remainders with each other 
and the product by half the sum of the sides, and the square 
root of this result is the area of the triangle. 

Example. 

Find the area of a triangle having sides 12 inches. 9 inches 
and 15 inches long. 

Solution : 

Half the sum of the sides = 18 

Area = \ 7 IS — 12) X .IS — 0; X (IS — 15; X IS 

Area = \A> X 9 X 3 X Is 

Area == \Z2916~ 

Area = 54 square inches. 




MENSURATION. 1 95 

To Find the Height in any Triangle when the Length 
of the Three Sides is Given. 

(See Fig. 2). 

The base line is to the sum of the other 
two sides as the difference of the sides is to 
the difference between the two parts of the 
base line, on each side of the line measuring 
the perpendicular height. If half this dif- 
ference is either added to or subtracted 
from half the base line, there will be obtained two right-angled 
triangles, in which the base and hypothenuse are known and 
the perpendicular may be calculated thus : Using Fig. 2 for an 
example, and adding half the difference to half the base line, 
this may be written in the formula: 

Rule. 

Multiply the sum of the sides by their difference and divide 
this product by twice the base ; to the quotient add half the 
base; square this sum (that is, multiply it by itself); subtract 
this from the square of the longest side, and the square root of 
the difference is the perpendicular height of the triangle. 

Example. 

In the triangle, Fig. 2, the sides are: 

c = 12 inches. 

a == 9 inches. 

b ■=. 6 inches ; find the perpendicular height x. 

x = J m _ / (9 + 6) X (9 - 6 ) 12V 
> " V 2X12 " r 2/ 

* = ^Sl-(li + 6) 2 
x = J 81 — 7.S75 2 
x = J 81 — 62.015 
x = J 18.985 
x = 4.357 inches. 



\2 



196 



MENSURATION. 



To Find the Area of a Parallelogram. 

Multiply the length by the width, and the product is the 



area. 



Note.— -The width must not be measured on the slant side, 
but perpendicular to its length. 

To Find the Area of a Trapezoid. 

Add the two parallel sides and divide by two ; multiply the 
quotient by the width, and the product is the area. (See Fig. 3). 





Fig. 3. 




k 


— 7 feet. • --»i 




jo/ 
1 / 






/ 
I \ 

b\ 




1 

i 


u 


9 feet. * 





Example. 

Find the area of 
(Fig. 3). 

Solution : 

7 + 9 



a trapezoid. 



Area 



X 4 = 32 square feet. 



Note. — The correctness of this may be best understood by 
assuming the triangle b cut off and placed in the position #, 
and the trapezoid will be changed into a rectangle 8 feet long 
and 4 feet wide. 

The area of any polygon may be found by dividing it into 
triangles and calculating the area of each separately, and the 
sum of the areas of all the triangles is the area of the polygon. 



The Area of a Circle. 

The area of a circle is equal to the square of the radius 
multiplied by 3.1416, which written in a formula is, 

Area = 3.1416 r 2 . 

The area of a circle is also equal to the 
square of the diameter multiplied by 0.7854, 
which may be written, 
Area = 0.7854 d 2 

The area of a circle is also equal to its 
circumference multiplied by the radius and 
the product divided by 2, which may be 
written, 

Area = C -*I 

2 

The correctness of these formulas may be best understood 
by assuming the circle to be divided into triangles (see Fig. 4), of 
which the height h = radius and the sum of the bases, b, of all 
the triangles is equal to the circumference of the circle. 




MENSURATION. 197 

Therefore, according to the formulas, 

.1 c , • 1 base X perpendicular height 

the area of a triangle = ±- — ±- ^ — 

& 2 

., r • 1 . u radius X circumference 
the area of a circle must be = ■ 

2 

and from this follow all the other formulas. 

To Change a Circle into a Square of the Same Area. 

Rule. 

Multiply the diameter of the circle by the constant 0.8862 
and the product is the length of one side in a square of the same 
area. 

Example. 

A circular water-tank 5 feet in diameter and 3 feet high is 
to be replaced by a square tank of the same height and volume. 
How long will each side in the new tank be? 

Solution : 

Side = 5 X 0.8862 = 4.431 feet long. 

To Find the Side of the Largest Square which can be 
Inscribed in a Circle. 

Rule. 

Multiply the diameter of the circle by the constant 0.7071 ;. 
the product is the length of the side of the square. 

Example. 

What is the largest square beam which can be cut from a 
log 30 inches in diameter. 

Solution : 

30 X 0.7071 = 21.213 inches square. 

Note. — A round log of any diameter will always cut into a 
square beam having sides seven-tenths the diameter of the 
round log. For instance, a 10-inch log will cut 7 inches square, 
a 15-inch log will cut 10.5 inches square, a 20-inch log will cut 
14 inches square, etc. 

To Find the Area of Any Irregular Figure. 

(See Fig. 5). 

Divide the figure into any 
number of equal parts, as shown 
by the perpendiculars 1, 2, 3, etc. 
Measure the width of the figure 
at the middle of each division; 
add these measurements together, 







Fig. 


5. 




i 
• 
1 

I 

1 


1 
1 
1 
1 

I 

1 


■ 1 

1 
1 
1 

1 

1 
1 

1 


1 
1 
1 
1 

1 


i 1 

1 ! 

1 

1 

1 

1 


1 


2 


3 


4 5 


6 


7 8 



I98 MENSURATION. 

divide this sum by the number of divisions (in Fig. 5 it is 8), 
multiply this quotient by the length a b, and the product is the 
area, approximately. 

Note. — Sometimes the figure is of such shape that it is 
more convenient to divide some of it into squares, rectangles, 
or triangles, and figure the rest as explained above. 



To Find the Area of a Sector of a Circle. 

The area of a sector of a circle is to the area of the whole 
circle as the number of degrees in the arc of the sector is to 360 
degrees. 

Thus: 

A _ r 2 X 3.1416 X a _ .008727 X r 2 X <z = — 
360 2 

r 2 A 



l= 3.1416 XaXr _ 0.0^45329 XaXr 



180 



r= 



a 



!J60A_ = ia7046 JA_ 
V 3.1416 a > a 

ISO/ 57.2956/ 



3.1416 r r 

A = Area of sector. 
r= radius of sector. 
a — number of degrees in arc. 
/= length of arc in same units as A and r. 

Example. 

The arc of the sector (Fig. 6) is 60° rie- 6. 

and the radius is 6 feet. Find area. 

360 __ 7T r 2 
60 Area 
60r 2 7v 



Area = 

Area = 



360 
60 X 6 X 6 X 3.1416 




360 
Area = 18.849 square feet. 

If the length of the arc is known instead of the number of 
degrees, multiply the length of the arc by the length of the 
radius, divide product by 2, and the quotient is the area of the 
sector. The correctness of this rule will be understood by the 
rule for area of circles, explained under Fig. 4. 



M ENSURATION. 1 9 9 

To Find the Length of Arc of a Segment of a Circle. 

The length of the arc may be calculated by the formula,* 

,-■ **- c 

o 
O 

/ = Length of arc, a f b \ 

c = Length of chord from a tof \ ( See Fig. 7). 
C = Length of chord from a to b ) 

Rule. 

Multiply the length of the chord of half the arc by 
8 ; from the product subtract the length of the chord of the arc ; 
divide the remainder by 3, and the quotient is the length of 
the arc. 

When chord and height of segment are known, the chord of 
half the arc is calculated thus : 

Chord of half the arc = /y/ n i _j_ ^2 

h = Height of segment (see d f, Fig. 7). 

n = Half the length of chord (see a d or b d, Fig. 7). 

When only the radius and the height of the segment are 
known, the length of the chord of the whole arc expressed in 
these terms will be : 2 X */ 2 r h h 2 

The chord of half the arc will be : ^2 r h 

Therefore the length of the arc will be : 

8 X V 2 r h — 2 X V '2 r h — h 2 
1 — 3 

/ = length of arc (a f b, Fig. 7 ). 
h = height of segment ( df, Fig. 7 ). 
r = radius of circle (cf, Fig. 7). 

To Find the Area of a Segment of a Circle. 

(See Fig. 7). 

Ascertain the area of the whole 
sector and from this area subtract the 
area of the triangle, and the rest is the 
area of the segment. 

Example. 

Find the the area of the segment 
when the radius is 9 inches and the 
arc 60°. 

* This formula is called " Huyghens's approximate formula for circular arcs," 
but it is so close that it may for any practical purpose be considered absolutely cor- 
rect for arcs having small center angles; for center angles as large as 120 ° , the 
lesult is only one quarter of one per cent, too small, and even for half a circle the 
result is scarcely more than one per cent, small as compared to results calculated by 
taking tt as 3.1416. 




200 MENSURATION. 

Solution : 

Area of segment — A— 60r2 * — 0.433 r 2 

360 

^ = 60X9X9X3.1416 _ 0433X9X9 

360 
A = 42.4116 — 35.073 
A = 7.3386 square inches. 

In this example the arc was 60°, consequently the triangle 
is equilateral ; therefore its area is found by the formula 
0.433 r 2 . ( See area of equilateral triangles, page 194). 

Note. — When the segment is greater than a semicircle, 
calculate by preceding rules and formulas the area of the lesser 
portion of the circle; subtract it from the area of the whole 
circle. The remainder is the area of the segment. 



To Find the Radius Corresponding to the Arc, when the 
Chord and the Height of the Segment Are Given. 

Rule. 

Add the square of the height to the square of half the 
chord ; divide this sum by twice the height, and the quotient is 
the radius. In a formula this may be written : 

' n* + & \ 

2h I 

I = radius = c b or cf \ ( See Fig. 7). 

n — half the chord = d b 1 

h — height = df ) 

The above rule and formula may be proved by rules for 
right-angled triangles; thus, c b or r equals hypothenuse, and ;/, 
or half the chord, equals perpendicular, and c d, which is equal 
tor — k, is the base. From the rule that the square of the 
hypothenuse is equal to the sum of the square of the base and 
the square of the perpendicular, we have : 

r 2 = n 2 + (r — h) 2 
r 2 = n 2 + r 2 — 2rh-\rh* 
r* — r 2 + 2 rh = n 2 + h 2 
2 r h = n 2 + h 2 
r - n 2 + h 2 
2k 

The perpendicular height of the triangle is always equal to 
the radius minus the height of the segment. ( See triangle a b c, 
and height, df, Fig. 7). 



MENSURATION. 



201 



TABLE No. 23. — Areas of Segments of a Circle. 

The diameter of a circle = 1, and it is divided into 100 
equal parts. 



h 


Area. 


h 


Area. 


h 


Area. 


D 




D 
0.18 




D 




0.01 


0.001329 


0.096135 


0.35 


0.244980 


0.02 


0.003749 


0.19 


0.103900 


0.36 


0.254551 


0.03 


0.006866 


0.20 


0.111824 


0.37 


0.264179 


0.04 


0.010538 


0.21 


0.119898 


0.38 


0.273861 


0.05 


0.014681 


0.22 


0.128114 


0.39 


0.283593 


0.06 


0.019239 


0.23 


0.136465 


0.40 


0.293370 


0.07 


0.024168 


0.24 


0. 144945 


0.41 


0.303187 


0.08 


0.029435 


0.25 


0.153546 


0.42 


0.313042 


0.09 


0.035012 


0.26 


0.162263 


0.43 


0.322928 


0.10 


0.040875 


0.27 


0.171090 


0.44 


0.332843 


0.11 


0.047006 


0.28 


0.180020 


0.45 


0.342783 


0.12 


0.053385 


0.29 


0.189048 


0.46 


0.352742 


0.13 


0.059999 


0.30 


0.198168 


0.47 


0.362717 


0.14 


0.066833 


0.31 


0.207376 


0.48 


0.372704 


0.15 


0.073875 


0.32 


0.216666 


0.49 


0.382700 


0.16 


0.081112 


0.33 


0.226034 


0.50 


0.392699 


0.17 


0.088536 


0.34 


0.235473 







Table No. 23 gives the areas of segments from 0.01 to 0.5 
in height when the diameter of the circle is 1. 

The area of any segment is computed by the following 
rule : 

Divide the height of the segment by the diameter of its 
corresponding circle. Find in the table in column marked 

—jy- the number which is nearest, and multiply the corresponding 
area by the square of the diameter of the circle, and the 
product is the area of the segment. 

Example. 

Figure the area of a segment of a circle, the height of the 
segment being 12 inches and the diameter of the circle 40 inches. 

Solution : 

12 divided by 40 = 0.3 

In the column marked -jy find 0.3 ; the corresponding 
area is 0.198168. 

The area of the segment is 40 X 40 X 0.198168 = 317.0688 
square inches, or 317 square inches. 



202 



MENSURATION. 



To Calculate the Number of Gallons of Oil in a Tank. 

Example. 

A gasoline tank car is standing on a horizontal track, and 
by putting a stick through its bung-hole on top it is ascertained 
that the gasoline stands 15 inches high in the tank. The 
diameter of the tank is 60 inches and the length is 25 feet. How 
many gallons of gasoline are there in the tank? 

Solution : 

15 divided by 60 is 0.25 

In Table Xo. 23, the area corresponding to 0.25 is 0.153546. 
Area of cross section of the gasoline is 60 X 60 X 0.153546 = 
552.7656 square inches. 

Twenty-five feet is 300 inches ; the tank contains 300 X 
552.7656 = 165829.68 cubic inches. One gallon is 231 cubic 
inches. The tank contains 165829.68 divided by 231 = 717.88, 
or 718 gallons. 

Note. — If the tank is more than half full, figure first the 
cubical contents of the whole tank if full, then figure the cubical 
contents of the empty space and subtract the last quantity from 
the first, and the difference is the cubical contents of the fluid 
in the tank. 

Circular Lune. 

The circular lune is a crescent-shaped 
figure bounded bv two arcs, as a b c and 
a dc. (Fig. 8). 

Its area is obtained by first finding 
the area of the segment a dc (having c<± 
for center of the circle), then the area 
of the segment a b c (having c\ for center 
of circle), then by subtracting the area of 
the last segment from the area of the first ; 
the difference is the area of the lune. 

A practical example of a circular lune is the area of the 
opening in a straight-way valve when it is partly shut. 




Circular Lune. 



Fig. 9. 




Circular Zone. 



Circular Zone. 

The shaded part, a b c d. of the figure 
is called a circular zone. Its area is ob- 
tained by first finding the area of the 
circle and then subtracting the area of the 
two segments ; the difference is the area of 
the zone. When the zone is narrow in 
proportion to the diameter, its area is ob- 
tained very nearly by following the rule : 
Add line a b or c d to the diameter of the 
circle, divide the sum by 2 and multiply 




MENSURATION. 203 

the quotient by the width of the zone, and the product is the 
area. 

To Compute the Volume of a Segment of a Sphere. 

Rule. fig. 

Square half the length of its base, and 
multiply by 3. To this product add square 
of the height. Multiply the sum by the 
height and by 0.5236. 

Example. 

Find volume of the spherical segment 
shown in Fig. 10 ; base line is 8" and 
height is 2". 

S olution : Segment of a Sphere. 

Volume = v = (3 X 4 2 + 2 2 ) X 2 X 0.5236 
v — (3 X 16 + 4) X 2 X 0.5236 
v — 52 X 2 X 0.5236 
v — 54.4544 cubic inches. 

To Find the Volume of a Spherical Segment, when the 

Height of the Segment and the Diameter of 

the Sphere are Known. 

Rule. 

Multiply the diameter of sphere by 3, and from this product 
subtract twice the height of segment. Multiply the remainder 
by the square of the height and the product by 0.5236. 

Example. 

The segment (Fig. 10) is cut from a sphere 10 inches in 
diameter and it is 2 inches high. Figure it by this last rule. 

Solution : 

Volume = v = (10 X 3 — 2 X 2) X 2 2 X 0.5236 
v = (30 — 4) X 4 X 0.5236 
v = 26 X 4 X 0.5236 
v = 54.4544 square inches. 

To Find the Surface of a Cylinder. 

Rule. 

Multiply the circumference by the length, and to this pro- 
duct add the area of the two, ends. 

A cylinder has the largest volume with the smallest surface, 
when length an4 cUarneter ar$ equal {o each other, 



204 



MENSURATION. 



To Find the Volume of a Cylinder. 

Rule. 

Multiply area of end by length of cylinder, and the product 
is the volume of the cylinder. 

Example. 

What is the volume of a cylinder -i inches in diameter and 9 
inches long? 

Solution : 

Area of end = r- ~ 

Volume = r- ~ 1—1 X 2 X 3.1416 X 9 = 113.0976 cubic inches. 



To Find the Solid Contents of a Hollow Cylinder. 

Rule. 

Find area of end according to outside diameter ; also find 
area according to inside diameter ; subtract the last area from 
the first and multiply the difference by the length of the 
cylinder. 

Formula : 

Area = {R* — r 2 ) - I 

R — Outside radius. 

r = Inside radius. 

/ = Length of cylinder. 

Example. 

Find the solid contents of a hollow cylinder of 6 feet outside 
diameter, 4 feet inside diameter and 5 feet long. 

Solution : 

Solid contents = x= (3 2 — 2 2 ) X 3.1416 X 5 

x — (9 — 4) X 3.1416 X 5 

x=b X 3.1416 X 5 

x = TS.54 cubic feet. 

To Find the Area of the Curved 
Surface of a Cone. 

(See Fig. 10). 

Rule. 

Multiply the circumference 
of the base' by the slant height 
and divide the product by 2 ; the 
quotient is the area of the curved 
surface. If the total surface is 
wanted, the area of the base is 
cone. added to the curved area. 







MENSURATION. 



20S 



If the perpendicular height is known, the length of the slant 
side or the slant height is found by adding the square of the per- 
pendicular height to the square of the radius and extracting the 
square root of the sum. 

Formula : 

Curved area = x = 5 — r n V r 2 + h* 

r = Radius of base. 
d= Diameter of base. 
h = Perpendicular height. 



To Find the Volume of a Cone. 



Rule. 



Multiply the area of the base by the perpendicular height, 
and divide the product by 3. 



By formula : 
Volume = 



T" 4 7T 



h 



To Find the Area of the Curved Surface of a Frustum 

of a Cone. 



(See Fig. 11). 

Rule. 

Add circumference of small end to 
circumference of large end, multi- 
ply this sum by the slant height and 
divide the product by 2. 

Formula : 

c* 
Curved area = (2 R-k 4- 2 r n) 

which reduces to 

Curved area = (R + r) ir S 

If the perpendicular height instead of 
the slant height is known, we have : 

Curved area = (R -j- r) -* \/ (ft rf + h? 

R = Large radius. 

r = Small radius. 

h = Perpendicular height, 

s = Slant height. 



Fi&fi. 




Frustum of a Cone. 



206 MENSURATION. 

To Find the Volume of a Frustum of a Cone. 

Rule. 

Square the largest radius ; square the smallest radius. 
Multiply largest radius by smallest radius ; add these three pro- 
ducts and multiply their sum by 3.1416; multiply this last 
product by one-third of the perpendicular height. 

Formula : 

Volume = Ctf 2 -f r 2 -f Rr) * — 

Example. 

Find the volume of a frustum of a cone. The largest 
diameter is 6 feet, the smallest diameter is 4 feet, and perpen- 
dicular height is 12 feet. 

Solution : 

Volume = x = (3 2 -f 2 2 + 3 X 2) X 3.1416 X — 

o 

x — (9 + 4 -f 6) X 3.1416 X 4 

x — 19 X 3.1416 X 4 

x = 238.7616 cubic feet. 

Note. — This rule will also apply for finding the solid con- 
tents of wood in a log. 

f.g. 12. To Find the Area of the S i an ted 

~kT \ Surface of a Pyramid. 

\V\ \ (See Fig. 12). 

/ / i \\ \ Rule. 

/ / i \ \ \ Multiply the length of the perim- 

/ / \ \ \ eter °^ tne Dase by the slant height of 

j /,-/- >- \-n\\ tne S ^e (not the slant height of the 

/'"'/ \ N >3^- e d§ e )- Divide the product by 2, and 

' ^^/ j \^ the quotient is the area. 

Pyramid 

To Find the Total Area of the Surface of a Pyramid. 

Rule. 

Find area of the slanted surface as explained above, and to 
this add the area of a polygon equal to the base of the pyramid. 

To Find the Volume of a Pyramid. 

Rule. 

Multiply the area of the base by one-third of the perpen- 
dicular height. 



MENSURATION. 



207 




Frustum of a Pyramid 



To Find the Area of the Slanted 

Surface of a Frustum of 

a Pyramid. 

Rule. 

Add perimeter of the small end 
to the perimeter of the large end. 
Multiply this sum by the slant height 
of the side (not slant height of edge). 
Divide the product by 2. 



To Find the Total Area of the Surface of a Frustum of a 

Pyramid. 

Rule. 

Find the area of the slanted surface as explained above, 
and to this area add the area of the two ends, 
obtained in the same way as areas of polygons. 



Their areas are 
(See page 196). 



To Compute the Volume of a Frustum of a Pyramid. 

Rule. 

Multiply the area of the small end by the area of the large 
end, extract the square root of the product, and to this add the 
area of the small end and the area of the large end; multiply the 
sum by one-third of the perpendicular height. 

Formula : 

Volume = A (a + A + V~A~a ) 

Example. 

Find volume of a frustum of a pyramid when the area of 
the small end is 8 square feet, the area of the large end is 18 
square feet and the perpendicular height is 30 feet. 

30 



Volume = v = — — y 
3 X 



v 



Rule. 



(8 + 18+Vl8X 8 ) 
10 X ( 8 4- 18 + V 144 ) 
v = 10 X ( 8 4- 13 + 12 ) 
v = 10 X 38 
v = 380 cubic feet. 

To Find the Surface of a Sphere. 



Multiply the circumference by the diameter. 

Note. — The surface of a sphere is equal to the curved sur- 
face of a cylinder having diameter and length equal to the 
diameter of the sphere. 



208 MENSURATION. 

To Find the Volume of a Sphere. 

Rule. 

Multiply the cube of the diameter by 3.1416, divide the 
product by 6 and the quotient is the volume of the sphere. Or, 
another rule is : Multiply the cube of the diameter by 0.5236 and 
the product is the volume of the sphere. 

Example. 

Find the volume of a sphere 15'' diameter. 

Solution : 

0.5236 X 15 X 15 X 15 = 1767.15 cubic inches. 

A sphere twice as large in diameter as another has twice the 
circumference, four times the surface, eight times the volume, 
and if of the same material will weigh eight times as much. 

To Compute the Diameter of a Sphere when the Volume 

is Known. 

Rule. 

Divide the volume by 0.5236 and the cube root of the 
quotient is the diameter of the sphere. 

To Compute the Circumference of an Ellipse. 

Rule. 

Add the square of the largest diameter to the square of the 
smallest diameter and divide the sum by 2 ; multiply the square 
root of the quotient by 3.1416. * 

Example. 

Find the circumference of an ellipse. The largest diameter 
is 24 inches and the smallest diameter is 18 inches. 

Solution : 



r- t omp /24 2 + 18 2 
Circumference = c = 3.1410 a 

H 2 

c = 3.1416 ^450 
c— 3.1416 X 21.2132 
c= 66.643 inches. 

To Compute the Area of an Ellipse. 

Rule. 

Multiply the smallest diameter by the largest diameter, and 
this product by 0.7854. 

* This Rule gives only approximate results. There is no known rule giving 
exact results. 



CIRCUMFERENCES AND AREAS OF CIRCLES. 



209 



TABLE No. 24. — Giving Circumferences and Areas of 

Circles. 













1 


Diameter. 


Circumfer- 


Area. 


Diameter. 


Circumfer- 


Area. 




ence. 






ence. 




■h 


0.0491 


0.00019 


43 
63 


2.1108 


0.35454 


X 


0.0982 


0.00077 


1 1 
1 6 


2.1598 


0.37122 


A 


0.1473 


0.00173 


4 5 

63 


2.2089 i 


0.38829 


1 


0.1964 


0.00307 


23 
3 2 


2.2580 


0.40574 


5 

63 


0.2454 


0.00479 


47 
6 4 


2.3071 


0.42357 


3 

3 2 


0.2945 


0.00690 


H 


2.3562 


0.44179 


7 
"63 


0.3436 


0.00940 


49 

64 


2.4053 


0.46039 


H 


0.3927 


0.01227 


25 
"3 2 


2.4544 


0.47937 


9 

63 


0.4418 


0.01553 


51 
64 


2.5035 


0.49874 


5 

3 2 


0.4909 


0.01918 


13 
16 


2.5525 


0.51849 


1 1 

63 


0.5400 


0.02320 


53 
64 


2.6016 


0.53862 


3 

1 6 


0.5890 


0.02761 


27 
3 2 


2.6507 


0.55914 


13 

63 


0.6381 


0.03241 


55 
6 4 


2.6998 


0.58004 


7 
3 2 


0.6872 


0.03758 


H 


2.7489 


0.60132 


1 5 
63= 


0.7363 


0.04314 


57 

6 4 


2.7980 


0.62299 


X 


0.7854 


0.04909 


29 
"3 2 


2.8471 


0.64504 


II 


0.8345 


0.05542 


5 9 

64 


2.8962 


0.66747 


9 

3 2 


0.8836 


0.06213 


15 
1 6 


2.9452 


0.69029 


19 
63 


0.9327 


0.06922 


61 
64 


2.9943 


0.71349 


5 
16 


0.9818 


0.07670 


3 1 
3 2 


3.0434 


0.73708 


21 


1.0308 


0.08456 


63 
6 4 


3.0925 


0.76105 


1 1 

3 2 


1.0799 


0.09281 


1 


3.1416 


0.78540 


23 

"c3 


1.1290 


0.10144 


h\ 


3.1907 


0.81013 


H 


1.1781 


0.11045 


I3V 


3.2398 


0.83525 


2 5 
6 4 


1.2272 


0.11984 


*A 


3.2889 


0.86075 


1 3 
3 2 


1.2763 


0.12962 


IrV 


3.3379 


0.88664 


2 7 
6 t 


1.3254 


0.13979 


!6 5 3 


3.3870 


0.91291 


7 
16 


1.3744 


0.15033 


I/2 


3.4361 


0.93956 


29 

"of 


1.4235 


0.16126 


IsV 


3.4852 


0.96660 


15 

3 2 


1.4726 


0.17258 


1/s 


3.5343 


0.99402 


3 I 

o3 


1.5217 


0.18427 


l- 9 - 

6 1 


3.5834 


1.02182 


^ 


1.5708 


0.19635 


1 5_ 

x 3 2 


3.6325 


1.05001 


3 3 

£ 


1.6199 


0.20881 


1 * 1 
1 6"3 


3.6816 


1.07858 


] 7 
3 2 


1.6690 


0.22166 


l- 3 - 

16 


3.7306 . 


1.10753 


3 5 
^3 


1.7181 


0.23489 


113 
x 64 


3.7797 


1.13687 


9 
T6 


1.7671 


0.24850 


l- 7 - 
3 2 


3.8288 


1.16659 


3 7 
63 


1.8162 


0.26250 


HI 


3.8779 


1.19670 


1 9 
"3"2 


1.8653 


0.27688 


IX 


3.9270 


1.22718 


3 9 


1.9144 


0.29165 


HI 


3.9761 


1.25806 


H 


1.9635 


0.30680 


i- 9 - 

x 3 2 


4.0252 


1.28931 


41 
63 


2.0126 


0.32233 


1 1 9 

- L 63 


4.0743 


1.32095 


21 
32 


2.0617 


0.33824 


J -l 6 


4.1233 


1.35297 



2IO 



CIRCUMFERENCES AND AREAS OF CIRCLES. 



Diameter. 


Circumfer- 
ence. 


Area. 


Diameter. 


Circumfer- 
ence. 


Area. 


1^1 


4.1724 


1.38538 


2% 


6.6759 


3.5466 


3 2 


4.2215 


1.41817 


2fV 


6.8722 


3. 7584 


Iff 


4. 2706 


1.45134 


2X 


7.0686 


3.9761 


in 


43197 


1.48489 


2A 


7.2649 


4.2 


125 

"4 


4.363S 


1.51883 


2H 


7.4613 


4.4301 


Hi 


4.4179 


1.55316 


*A 


7.6576 


4.6664 


lit 


4.4670 


1.58786 


2% 


7.8540 


4.90S7 


i^ 


4.5160 


1.62295 


2x 9 i 


8.0503 


5.1573 


i : - 

A 6 4 


4.5651 


1.65843 


2H 


8.2467 


5.4119 


115. 
-••3 2 


4.6142 


1.69428 


2H 


8.4430 


5.6727 


m 


4.6633 


1.73052 


2% 


S.6394 


5.9396 


i% 


4.7124 


1.76715 


Ol 3 

S T6 


B.8357 


6.2126 


in 


4.7615 


1.80415 


2-/g 


9.0321 


6.4918 


Hi 


4.8106 


1.84154 




9.22-4 


6.7772 


13 5 

A 6 4 


4.8597 


1.87932 


3 


9.4248 


7.0686 


- L 16 


4.9087 


1.91748 


3 T V 


9.6211 


7.3662 


1§I 

J ->5 4 


4.9578 


1.95602 


sy s 


9.8175 


7.6699 


119 

1 32 


5.0069 


1.99494 


»A 


10.0138 


7.9798 


139 

1 64 


5.0560 


2.03425 


3X 


10.2102 


8.2958 


1# 


5.1051 


2.07394 


O 5 
'-> 1 6 


10.4066 


8.6179 


Ifl 


5.1542 


2.11402 


m 


10.6029 


8.9462 


121 
A 3 2 


5.2033 


2.15448 


3 T V 


10.7992 


9.2807 


x 64 


5.2524 


2.19532 


3^ 


10.9956 


9.6211 


111 
X 1S 


5.3014 


2.23654 


3 r 9 ,- 


11.1919 


9.9678 


lit 


5.3505 


2.27815 


m 


11.3883 


10.3206 


123 

1 3 2 


5.3996 


2.32015 


3|| 


11.5846 


10.6796 


lit 


5.44-7 


2.36252 


3K 


11.7810 


11.0447 


IK 


5.4978 


2.40528 


Bfl 


11.9773 


11.4160 


119 
x 64 


5.5469 


2.44843 


3^ 


12.1737 


11.7933 


115 

x 3 2 


5.5960 


2.49195 


915 

°T6 


12.3701 


12.1768 


151 
1 54 


5.6450 


2.53586 


4 


12.5664 


12.5664 


lia 

x l 6 


5.6941 


2.58016 


4 T V 


12.7628 


12.9622 


153 

± 64 


5.7432 


2.62483 


4>i 


12.9591 


13.3641 


III 

*3 3 


5.7923 


2.66989 


*r 3 s 


13.1554 


13.7721 


15 5 
x 64 


5.8414 


2.71534 


4X 


13.3518 


14.1863 


1# 


5.8905 


2.76117 


4fV 


13.5481 


146066 


lit 


5.9396 


2.80738 


4^ 


13.7445 


15.0330 


129 
-3 2 


5.9881 


2.85397 


*i^ 


13.9408 


15.4656 


*64 


6.0377 


2.90095 


±Yz 


14.1372 


15.9043 


111 
- 1 I tj 


6.0868 


2.94831 


*r 9 s 


14.3335 ! 


16.3492 


*e 4 


6.1359 


2.99606 


4^ 


14.5299 


16.8002 


131 
J 3 2 


6.1850 


3.04418 


4H 


14.7262 


17.2573 


16 3 
-"■6 4 


6.2341 


3.0927 


4* 


14.9226 


17.7206 


2 


6.2832 


3.1416 


4.13 

* 1 >5 


15.1189 ! 


18.19 


9 1 


6.4795 


3.3410 


4^ 


15.3153 


18.6655 



CIRCUMFERENCES AND AREAS OF CIRCLES. 



211 



Diameter. 


Circumfer- 
ence. 


Area. 


Diameter. 


Circumfer- 
ence. 


Area. 


415 


15.5116 


19.1472 


10^ 


32.9868 


86.5903 


5 


15.7080 


19.6350 


10y 8 


33.3795 


88.6643 


5^ 


16.1007 


20.6290 


iou 


33.7722 


90.7625 


b% 


16.4934 


21.6476 


ioy 8 


34.1649 


92.8858 


&H 


16.8861 


22.6907 


ii 


34.5576 


95.0334 


$Yz 


17.2788 


23.7583 


ny* 


34.9503 


97.2055 


z% 


17.6715 


24.8505 


nx 


35.343 


99.4019 


&K 


18.0642 


25.9673 


UH 


35.7357 


101.6234 


5^ 


18.4569 


27.1084 


ny 2 


36.1284 


103.8691 


6 


18.8496 


28.2744 


nn 


36.5211 


106.1394 


§A 


19.2423 


29.4648 


HX 


36.9138 


108.4338 


6X 


19.635 


30.6797 


liH 


37.3065 


110.7537 


QH 


20.0277 


31.9191 


12 


37.6992 


113.098 


6/ 2 


20.4204 


33.1831 


12X 


38.4846 


117.859 


QH 


20.8131 


34.4717 


12)4 


39.2700 


122.719 


6U 


21.2058 


35.7848 


12% 


40.0554 


127.677 


en 


21.5985 


37.1224 


13 


40.8408 


132.733 


7 


21.9912 


38.4846 


13X 


41.6262 


137.887 


^A 


22.3839 


39.8713 


n x A 


42.4116 


143.139 


?X 


22.7766 


41.2826 


im 


43.1970 


148.490 


in 


23.1693 


42.7184 


14 


43.9824 


153.938 


W 


23.5620 


44.1787 


14X 


44.7678 


159.485 


ny, 


23.9547 


45.6636 


14^ 


45.5532 


165.130 


IK 


24.3474 


47.1731 


U% 


46.3386 


170.874 


7% 


24.7401 


48.7071 


15 


47.1240 


176.715 


8 


25.1328 


50.2656 


15X 


47 9094 


182.655 


8/s 


25.5255 


51.8487 


15^ 


48.6948 


188.692 


8% 


25.9182 


53.4561 


Vo% 


49.4802 


194.828 


&A 


26.3109 


55.0884 


16 


50.2656 


201.062 


8/ 2 


26.7036 


56.7451 


16X 


51.051 


207.395 


8% 


27.0963 


58.4264 


16K 


51.8364 


213.825 


8U 


27.489 


60.1319 


1Q% 


52.6218 


220.354 


8ft 


27.8817 


61.8625 


17 


53.4072 


226.981 


9 


28.2744 


63.6174 


nx 


54.1926 


233.706 


$A 


28.6671 


65.3968 


vjy 2 


54.9780 


240.529 


9X 


29.0598 


67.2008 


n# 


55.7634 


247.450 


QH 


29.4525 


69.0293 


18 


56.5488 


254.470 


vy 2 


29.8452 


70.8823 


18X 


57.3342 


261.587 


9tt 


30.2379 


72.7599 


18^ 


58.1196 


268.803 


9U 


30.6306 


74.6619 


18X 


58.905 


276.117 


9Vs 


31.0233 


76.5888 


19 


59.6904 


283.529 


10 


31.4160 


78.5400 


19X 


60.4758 


291.040 


ioy 8 


31.8087 


80.5158 


19^ 


61.2612 


298.648 


iox 


32.2014 


82.5158 


19X 


62.0466 


306.355 


ion 


32.5941 


84.5409 


20 


62.8320 


314.16 



212 



CIRCUMFERENCES AND AREAS OF CIRCLES. 



Diameter. 


Circumfer- 
ence. 


Area. 


Diameter. 


Circumfer- 
ence. 


Area. 


21 


65.9736 


346.361 


66 


207.34 


3421.19 


22 


69.1152 


380.134 


67 


210.49 


3525.65 


23 


72.2568 


415.477 


68 


213.63 


3631.68 


24 


75.3984 


452.39 


69 


216.77 


3739.28 


25 


78.540 


490.87 


70 


219.91 


3848.45 


26 


81.681 


530.93 


71 


223.05 


3959.19 


27 


84.823 


572.56 


72 


226.19 


4071.50 


28 


87.965 


615.75 


73 


229.34 


4185.39 


29 


91.106 


660.52 


74 


232.48 


4300.84 


30 


94.248 


706.86 


75 


235.62 


4417.86 


31 


97.389 


754.77 


76 


238.76 


4536.46 


32 


100.53 


804.25 


77 


241.90 


4656.63 


33 


103.67 


855.30 


78 


245.04 


4778.36 


34 


106.81 


907.92 


79 


248.19 


4901.67 


35 


109.96 


962.11 


80 


251.33 


5026.55 


36 


113.10 


1017.88 


81 


254.47 


5153.00 


37 


116.24 


1075.21 


82 


257.61 


5281.02 


38 


119.38 


1134.11 


83 


260.75 


5410.61 


39 


122.52 


1194.59 


84 


263.89 


5541.77 


40 


125.66 


1256.64 


85 


267.04 


5674.50 


41 


128.81 


1320.25 


80 


270.18 


5808.80 


42 


131.95 


1385.44 


87 


273.32 


5944.68 


43 


135.09 


1452.20 


88 


276.46 


6082.12 


44 


138.23 


1520.53 


89 


279.60 


6221.14 


45 


141.37 


1590.43 


90 


282.74 


6361.73 


46 


144.51 


1661.90 


91 


285.88 


6503.88 


47 


147.65 


1734.94 


92 


289.03 


6647.61 


48 


150.80 


1809.56 


93 


292.17 


6792.91 


49 


153.94 


1885.74 


94 


295.31 


6939.78 


50 


157.08 


1963.50 


95 


298.45 


7088.22 


51 


160.22 


2042.82 


96 


301.59 


7238.23 


52 


163.36 


2123.72 


97 


304.73 


7389.81 


53 


166.50 


2206.18 


98 


307.88 


7542.96 


54 


169.65 


2290.22 


99 


311.02 


7697.69 


55 


172.79 


2375.83 


100 


314.16 


7853.98 


56 


175.93 


2463.01 


101 


317.30 


8011.85 


57 


179.07 


2551.76 


102 


320.44 


8171.28 


58 


182.21 


2642.08 


103 


323.58 


8332.29 


59 


185.35 


2733.97 


104 


326.73 


8494.87 


60 


188.50 


2827.43 


105 


329.87 


8659.01 


61 


191.64 


2922.47 


106 


333.01 


8824.73 


62 


194.78 


3019.07 


107 


336.15 


8992.02 


63 


197.92 


3117.25 


108 


339.29 


9160.88 


64 


201.06 


3216.99 


109 


342.43 


9331.32 


65 


204.20 


3318.31 


110 


345.58 


9503.32 



Strength of flfeatedals. 



The strength of materials may be divided into Tensile, 
Crushing, Transverse, Torsional, or Shearing, and besides 
this, the elasticity of the material or its resistance against 
deflection must also be taken into consideration in figuring for 
strength. 

Tensile Strength. 

From experiments it is known that it it will take from 
40,000 to 70,000 pounds to tear off a bar of wrought iron one 
inch square. Therefore we usually say that the tensile strength 
of wrought iron is from 40,000 to 70,000 pounds, according to 
quality. The average is 50,000 to 55,000 pounds. The tensile 
strength of any body is in proportion to its cross sectional area ; 
thus, if a bar of iron of one square inch area will pull asunder 
under a load of 40,000 pounds, it will take 80,000 pounds to pull 
asunder another bar of the same kind of iron but of two square 
inches area. The tensile strength is independent of the length 
of the bar, if it is not so long that its own weight must be 
taken into consideration. Table No. 24 gives the load which 
will pull asunder one square inch of the most common materials. 

No part of any machine should be strained to that limit. A 
high factor of safety must be used, sometimes from 4 to 30 or 
even more, which will depend upon the kind of stress the mem- 
ber is exposed to, as dead load, variable load, shocks, etc. Dif- 
ferent factors of safety are also used for different kinds of 
material. (See page 274). 

ilodulus of Elasticity. 

The modulus of elasticity for any kind of material is usually 
defined as the amount of force which would be required to 
stretch a straight bar of one square inch area to double its 
length or compress it to nothing, if this were possible. But a 
more comprehensive definition is to say that the modulus of 
elasticity is the reciprocal of the fractional part of the length 
which one unit of force will, within elastic limit, stretch or com- 
press one unit of area. For instance, if the modulus of elasticity 
for a certain kind of wrought iron is 25,000,000, it means that it 
would take 25,000,000 pounds of pulling force to stretch a bar of 

(213) 



214 STRENGTH OF MATERIALS. 

one square inch area to double its length, if this could possibly 
be done ; but it means also — which is exactly equivalent — 
that one pound of pulling force will stretch a bar of one square 
inch area one 25-millionth part of its length, or one pound 
compressive force will shorten the same bar one 25-millionth 
part of its length, and that two pounds of force will stretch or 
compress twice as much, three pounds thrice as much, etc. 

Strength of Wrought Iron. 

From experiments it is known that wrought iron can not 
very well be stretched or compressed more than one-thousandth 
part of its length without destroying its elasticity ; therefore if 
a bar of wrought iron has 25,000,000 as its modulus of elas- 
ticity, one pound will stretch it ^5 oV 00 of its length and it 
would take 25,000 pounds to stretch it toVo of its length. Thus, 
25,000 pounds would then be said to be its strength at the limit 
of elasticity for that kind of iron ; 80 to 100 per cent, more will 
usually be the ultimate breaking load. 

The pull or load which such a bar can sustain with safety 
will depend a great deal on circumstances, but it must never 
exceed 25,000 pounds per square inch of area. It must not 
even approach this limit if the structure is of any importance or 
if the load is to be sustained for any length of time, or if it is, 
besides the load, also exposed to shocks or jar. 

Strength of Cast Iron. 

Cast iron of good quality has a modulus of elasticity of 
15,000,000 pounds, but if strained so it will stretch ysV of its 
length its elasticity is usually destroyed. For instance, a bar of 
cast iron of one square inch area is exposed to tensile strain, 
its modulus of elasticity being 15,000,000 pounds and its elas- 
ticity being destroyed if it stretches T5V0 of its length, what 
then would be its strength at limit of elasticity? One pound 
will stretch it T ___i -^ of its length, therefore it must take 
10,000 pounds to stretch it y^Vo of its length; thus we would 
say that 10,000 pounds is its strength at limit of elasticity. It 
is not always that cast iron is of as good quality as that; very 
frequently its elasticity is destroyed if it is exposed to a tensile 
stress of 6,000 pounds per square inch of area; thus the strength 
of cast iron at its limit of elasticity is often found to be only 
6,000 pounds instead of 10,000 pounds. Besides, it is very 
often found that a pulling force of 10,000 pounds will stretch 
a bar of one square inch area one twelve-hundredth part of its 
length, and this, of course, gives the modulus of elasticity 
12,000,000 pounds. Frequently cast iron is of such quality that 
it cannot be stretched over o 5V of its len g tn before its elas- 
ticity is destroyed. Cast iron is very variable in quality, and 



STRENGTH OF MATERIALS. 21 5 

especially so with regard to its tensile strength. Generally 
speaking, we may say that for cast iron the 

Modulus of elasticity is 12,000,000 to 15,000,000 pounds. 

Tensile strength at limit of elasticity, 5,000 to 10,000 pounds. 

Ultimate tensile strength, 10,000 to 20,000 pounds. 

Elongation Under Tension. 

The total stretch or elongation of any specimen when 
exposed to tensile stress within the elastic limit is directly pro- 
portional to the length of the specimen, but it is inversely 
proportional to the modulus of elasticity and the cross sectional 
area of the specimen. The following formulas may, therefore, 
be used in such calculations : 

E _P X L p — E x s x A 

s X A L 

s - FX L A- PXL 

E X A s XE 

L — E X sX A 

P 

E ■=■ Modulus of elasticity in pounds per square inch. 
P = Load or force in pounds acting to elongate the specimen. 
s = Total stretch of specimen in inches in the length L. 
L = Original length of specimen in inches before force is 

applied. 
A = Cross-sectional area of specimen in square inches. 

Example. 

From experiments it is known that the modulus of elasticity 
for a certain kind of wrought iron is 28,000,000 ; what will then 
be the total stretch or elongation in a round boiler stay, IX 
inches in diameter and 6 feet long, when exposed to a stress of 
5000 pounds? 

Solution : 

1 % inches diameter = 1.227 inches area (see table, page 209) 
6 feet long = 72 inches. 

P X L 

E X A 

_ 5000 X 72 

28000000 X 1227 
s = 0.0105 inches = total stretch in the stay. 

Note. — As already stated, wrought iron can not be 
stretched as much as one-thousandth part of its original length 
without danger of destroying its elasticity; thus, for this stay, 
which is 72 inches, the limit of elasticity will be at a stretch of 
0.072 inches ; therefore the stretch produced by a load of 5,000 
pounds, which is calculated to be 0.0105 inches, is well within 
the safe limit. 



2l6 



STRENGTH OF MATERIALS. 



TABLE No. 25.— Modulus of Elasticity and Ultimate 
Tensile Strength of Various Materials. 







Ultimate 


Modulus of 


Ultimate 




Modulus of 


Tensile 


Elasticity in 


Tensile 


Materials. 


Elasticity in 
Pounds per 


Strength in 
Pounds per 


Kilograms per 
Square 


Strength in 
Kilograms 




Square Inch. 


Square Inch. 


Centimeter. 


per Square 
Centimeter. 


Cast steel . . . 


30.000.000 


100,000 


2,200.000 


7,000 


Bessemer steel . . 


28.000,000 


70.000 


1.970,000 


4.930 


Wrought iron bars 


.25,000.000 


55,000 


1,700.000 


3,850 


Wrought iron wire 


28,000.000 


75,000 


1,970,000 


5,250 


( 


12,000.000 


10,000 


80,0000 


700 


Cast iron * . . . j 


to 


to 


to 


to 


( 


15,000,000 


20,000 


1.000.000 


1,400 


Copper bolts . . . 


18.000,000 


35.000 


1.200,000 


2,400 


Brass . . . 






9,000,000 


17,700 


630,000 


1,200 


Oak . . . 








1,500,000 


17,000 


105,000 


1,200 


Hickory- 








1.400,000 


20,000 


98,000 


1,400 


Maple . . 








1,100,000 


15,000 


77,000 


1.000 


Pitch pine 








1,600,000 


15,000 


112,000 


1,000 


Pine . . 








1,100,000 


10,000 


77,000 


700 


Spruce . 








1,100,000 


10.000 


77.000 


700 



The two last columns in above table are calculated by the 
rule : One pound per sq. inch = 0.07031 kilograms pei sq. centi- 
meter and the result is reduced to the nearest round number. 



Formulas for Tensile Strength. 

The ultimate tensile strength of any specimen is in propor- 
tion to its cross-sectional area, and is expressed by the following 
formula : 



p = 


AX S 


s = 


P 
A 


A = 


P 

S 



Side of a square bar = -v 



fF 



S 



Diameter of a round bar 



4 



p 



S X 0.7854 



P r= Force in pounds which will pull the specimen asunder. 

S = Ultimate tensile strength in pounds per square inch. 
(See Table No. 25). 

A = Cross-sectional area of the specimen in square inches. 



* Very strong cast iron may have an ultimate tensile strength as high as 30,000 
pounds per square inch. 



STRENGTH OF MATERIALS. 2 I 'J 

Example. 

A piece of iron y 2 inch square is tested in a testing machine 
and breaks at a total stress of 14.210 pounds. What is the 
ultimate tensile strength per square inch ? 

Solution : 

A bar y> inch square has a cross-sectional area of y X y n 
is X square inch. 

S = — — = 56,840 pounds per square inch. 

Example. 

What will be the breaking load for a wrought iron bar 
H" X y%" when exposed to tensile stress, the ultimate tensile 
strength of the iron being 55,000 pounds per square inch, as 
given in Table No. 25, page 216? 

Solution : 

A bar y X y," is / T square inches in area. 

P =■ 6 9 ¥ X 55,000 = 7734 pounds, which will break the bar. 

In order to obtain the safe working stress introduce a suit- 
able factor of safety, from 5 to 10, according to circumstances, 
and calculate by the following formulas : 

PXf=AXS \~PX~f 
a s/ <r Side of a square bar = -v/ -L 



P — 



T 



a P X f Diameter of a round bar = \ J— 

A 1^ > 0.7854 S 

P = Load in pounds. 
/= Factor of safety. 

Example. 

A load of 24,000 pounds is suspended on a round wrought 
iron bar. The ultimate tensile strength of the iron is 55,000 
pounds per square inch. What should be the diameter of the 
bar to sustain the load, with 10 as the factor of safety ? 

Solution : 

S 

A = 240Q0 X ]0 = 4.363 square inches. 
55000 

In Table No. 24, we find the nearest larger diameter to be 
2y inches. 

The diameter may also be calculated directly by the fol- 
lowing formula : 



2lS STRENGTH OF MATERIALS. 



*S X 0.7854 



D _ / 24000 X 10 
' > 55000 X 0.785- 



1854 

D = s/ 5.56 

D = 2.358, or nearly 2?& inches diameter. 

To Find the Diameter of a Bolt to Resist a Given Load. 

Rule. 

Multiply pull in pounds by the factor of safety. Multiply 
the ultimate tensile strength of the material by 0.7854 : divide 
this first product by the last and extract the square root from 
the quotient which will then be diameter of bolt at the bottom 
of the thread. 

P = 
f= 



S X 0.7854 
D 2 X S X 0.7854 

f 
n z X S X 0.7854 



P 

D = Diameter of bolt or screw in the bottom of the thread. 
P = Load or pull in pounds. 
f== Factor of safety. 

S = Ultimate tensile strength per square inch. 
0.7854 is constant = __ 

Note. — Bolts are frequently exposed to a considerable 
amount of initial stress, due to the tightening of nuts, which 
must always be allowed for when deciding upon the load to be 
considered when calculating their diameter. 

Example. 

Find diameter of a bolt to sustain a load of 4.450 pounds, 
taking 10 as factor of safety and ultimate tensile strength of the 
iron to be 50.000 pounds per square inch. 

Solution : 

D — \i 445 ° X 10 

^50000 X 0.7854 

D = 1.064" in the bottom of thread ; thus, a 1 %" screw, 
standard thread, which is lj 1 - inches in diameter at the bottom 
of thread, will be the bolt to use. 



STRENGTH OF MATERIALS. 2IQ 

Example 2. 

What size of bolt is required to sustain the same load as is 
mentioned in the previous example, if only 5 is wanted as a 
factor of safety? 

Solution : 



p __ I 4450 X 5 

" * 50000 X 0.7854 

D — \A).567 

D = 0.75 inch diameter in bottom of thread. 

Thus a J^-inch standard screw is too small, as that is only 
f|" in bottom of thread, but a 1-inch standard screw is suffi- 
cient, being §£" in bottom of thread. 

To Find the Thickness of a Cylinder to Resist a Given 

Pressure. 

When the walls of cylinders are thin in proportion to their 
diameters use the formula : 

p = S X / 



R x/ 

f _ P X RXf 



R = 



S 
S X / 



P Xf 

P = Pressure per square inch. 
R = Radius of cylinder in inches. 

t = Thickness of cylinder wall in inches. 
/■= Factor of safety. 

S = Ultimate tensile strength of material. 

When cylinder walls are thick in proportion to the diameter, 
such as hydraulic cylinders, their thickness is usually figured 
by the formula : 

P X R 

t — Thickness of cylinder wall in inches. 
P = Pressure in pounds per square inch. 
R = Radius of cylinder. 

S = Ultimate tensile strength. 

f= Factor of safety. 
Example. 

Find necessary thickness of a hydraulic cylinder of 10-inch 
inside diameter, made from cast-iron,'to stand a pressure of 1000 



2 20 STRENGTH OF MATERIALS. 

pounds per square inch, with 4 as factor of safetv. The ulti- 
mate tensile strength of the iron is. by experiments." found to be 
20,000 pounds per square inch. ( See Table No. 25 '). 
Solution : 

10-inch diameter = 5-inch radius. 
1000 X 5 



t — 



20000 

—4— — 1000 



t __ 1000 X 5 



/ = 



5000 — 1000 
5000 



4000 
f = l% inch. 

Strength of Flat Cylinder Heads. 

The American Machinist, in Question No= 147, March 22, 
1894, gives the following formula for flat circular heads firmly 
fixed to the flange of the cylinder : 

f — J 2 X f 2 X P 
> 3 X .Si 

/ = Thickness of cylinder head in inches, 
r = Radius of cylinder head in inches. 
P =■ Pressure in pounds per square inch. 
Si =■ Allowable working stress in the material. 

The allowable working stress may be taken as ■§■ to ^ of 
the ultimate tensile strength and may. for cast iron, be from 
1500 to 2500 and for wrought iron from 4000 to 6000. The 
above formula was used to calculate the thickness of a cast iron 
cylinder head of 30 inches diameter, to resist a pressure of 100 
pounds per square inch. This formula is in that case con- 
sidered to give sufficient thickness, so that no ribs or braces are 
needed. 

The above formula may also be used for wrought iron, by 
selecting the proper value for Si. Assuming the tensile strength 
of wrought iron to be 44.000 pounds, and allowing a factor of 
safety of 8, the value of Si for wrought iron will be 5500. 

Strength of Dished Cylinder Heads. 

The American Machinist, in Question 1S3, April 12, 1894, 
gives the following formula for dished circular heads, firmly 
fixed to the flanges of the cylinder : 

f - PX {R* + d*) 
4 X Si X d 



STRENGTH OF MATERIALS. 22 1 

/ = Thickness of cylinder head in inches. 
R = Radius of cylinder head in inches. 
P ■=. Pressure in pounds per square inch. 

d = Depth in inches of dishing of the head at its center. 
■5"i = Allowable working stress in the material, which may 
be the same as given above. 

This formula was used to calculate the thickness of a cast 
iron head 44 inches in diameter, dished 7 inches, steam pressure 
75 pounds per square inch. 

Note. — In these examples the radius of the bolt circle 
should be considered as the radius of the head when calculat- 
ing the thickness. For diameter of bolts and spacing of bolts 
see examples under Steam Engine. 

The above formula may also be used for dished cylinder 
heads of wrought iron or steel, by allowing the proper value for 
•5*1. For soft steel Si may be 9000 to 12,000 pounds, and for 
wrought iron 5000 to 8000 pounds. 

Caution. — Cast iron is not a desirable material to use for 
large unribbed cylinder heads; either flat or dished wrought 
iron or steel is far superior. 

Strength of a Hollow Sphere Exposed to Internal 

Pressure. 

The pressure acts on a surface equal to — ~ A — and it is re- 

7~) | sf 

sisted by a metal area equal to — ~ — X ^ X t. 

D = External diameter. 
d = Internal diameter. 
/ = Thickness of metal. 

When the difference between inside and outside diameter 
is small it need not be considered in practice, and the formula 
will be : 

d 2 ~ 
P X — £- = dX * X t X Si 

which reduces to 

p _ 4 X / X S i 
d 
P X d 
*— 4: Si 
Si = Allowable tensile stress in the material. 
Note. — This formula only allows for tensile strength; if it 
is used for calculating the thickness of the body of a globe 
valve or anything similar a liberal amount of metal must be, 
added, in order to obtain good results when casting. 



222 



STRENGTH OF MATERIALS. 



Strength of Chains. 

The following table gives approximately the weight of 
wrought iron chains, in pounds per foot and kilograms per 
meter; and also their strength, with six as factor of safety. 
Chains ought to be tested with twice the load given in the 
table. Never lose sight of the fact that a chain in use will wear 
and consequently become reduced in strength; also, that a 
chain is no stronger than its weakest link. 



Diameter 
of Links 
in Inches. 


Load 
in Pounds. 


Weight in 

Pounds 
Per Foot. 


Load 
in Kilograms. 


Weight in Kilo- 
grams per Meter. 


3 
1 6 


280 


0.42 


125 


0.625 


X 


500 


0.91 


225 


1.35 


H 


1125 


1.5 


510 


2.22 


% 


2000 


2.5 


900 


3.72 


Va 


4500 


5.8 


3050 


8.63 


l 


8000 


10 


3600 


14.88 



Strength of Iron Wire Rope. 

The following table gives approximately the strength of 
iron wire rope, with six as a factor of safety. 



Diameter in Inches. 


Load in Pounds. 


Load in Kilograms. 




1000 
2500 
3500 


453 
1134 

1588 



Wire ropes should not be bent over pulleys of very small 
diameter. When used for hoisting, the diameter of pulley ought 
at least to be 40 times the diameter of rope. For further 
information on wire rope see manufacturers' catalogues. 

Strength of Manila Rope. 

The size of manila rope is measured by the circumference, 
therefore so-called three-inch rope is about one inch in diameter. 
New manila rope of three inches circumference will usually 
break for a load of 7,000 to 9,000 pounds. For common use 
such ropes may be loaded as given in the following table, and 
the diameter of the pulley ought to be at least eight times the 
diameter of the rope. 



Size of Rope, 


Safe Load in Pounds. \ Safe Load in Kilograms. 


3 ins. circumference, 
5 " " 


500 

800 

1300 


227 
363 
590 



STRENGTH OF MATERIALS. 



223 



CRUSHING STRENGTH. 

Short posts having square ends well fitted may be 
considered to give away under pure crushing stress. Their 
strength is in proportion to their area; therefore, when the 
length of a post does not exceed four to five times its diameter 
or smallest side, its strength or size may be calculated by the 
following formulas : 



P = 



A X S 



f 
S 



Side of a square post 



Diameter of a round post 



=v 



PXf 

s 



= V^ 



x/ 



(854^ 

P — Safe load in pounds to be supported by the post. 

A == Area of post in square inches. 

S = Ultimate crushing strength of the material in pounds 
per square inch, given in Table No. 26. 

f = Factor of safety. 

TABLE No. 26. — Modulus of Elasticity and Ultimate 
Crushing Strength of Various Materials. 





Modulus of 


Ultimate 


Modulus of 


Ultimate 




Elasticity 


Crushing 


Elasticity in 


Crushing 


Materials. 


in Pounds 


Strength in 


Kilograms 


Strength in 




per Square 


Pounds per 


per Square 


Kilograms per 




Inch. 


Square Inch 


Centimeter. 


Sq. Centimeter 






150,000 


2,200,000 


10,500 


Bessemer steel . . 


28,000,000 


50,000 
45,000 


1,970,000 


3,500 
3,000 


Wrought iron . < 


25,000,000 
12,000,000 


to 
50,000 


1,700,000 
800,000 


to 
3,500 


Cast iron . . . < 


to 
15,000,000 


90,000 


to 
1,000,000 


6,300 


Oak (endwise) 


1,500,000 


9,000 


105,000 


630 


Pitch pine " 


1,600,000 


9,000 


112,000 


630 


Pine 


1,100,000 


6,000 


77,000 


420 


Spruce " 


11,000,000 


6,000 


77,000 


420 


Brick | 




800 
to 2,000 


.... 


56 
to 140 


Brick work laid ) 










inl part cement > 




600 


.... 


42 


and 3 parts sand ) 










Brick work laid ) 
in lime and sand J 





240 


. • . 


16 to 17 










Granite 





10,000 


.... 


700 



224 



STRENGTH OF MATERIALS. 



Fig. 



When a post or column is long compared to its diameter, 
its strength will decrease as the length is increased. Anyone 
will, from every-day observation, know that a short post will 
support with perfect safety a load which will break a long one. 

Short columns break under crushing, but long ones break 
under comparatively light load by the combined effect due to 
both crushing and flexure. It is, therefore, evident that the 
strength of long columns follows laws very different from those 
which apply to short ones. 

The form of the ends has also 
great influence on the strength of a 
column when under crushing and 
deflective stress. ( See Fig. 1). 

When both ends are round the 
column has least strength; if one 
end is round and one end flat it is 
stronger, but if both ends are flat 
and square with the center-line, it is 
strongest. The proportions are ap- 
proximately as 1, 2 and 3. 

Eccentric loading on columns will also have a very destruct- 
ive effect upon their strength. 

Theoretical calculations regarding the strength of columns 
and posts are difficult, and such empirical formulas as the 
well-known Hodgkingson's or Gordon's formulas are usually re- 
sorted to. 

The Hodgkingson formulas for long columns having square 
ends well fitted are : 



^zd^y 



TZZZZZ? 



Z&U&d*' 



P = 99,000 X 



J) 3-55 
X 1 ^ 



for solid cast iron columns. 



r)3.5o d 3-55 

p = 99,000 X for hollow cast iron columns. 

£1-7 



P— 285,000 X 



£>3.5 



for solid wrought iron columns. 



P = Breaking load in pounds. 
D= External diameter in inches. 
d= Internal diameter in inches. 
L = Length in feet. 

When the breaking load as calculated by these formulas 
exceeds one quarter of the crushing load of a short column of 
the same metal area, the result must be corrected by the for- 
mula : 



Pl = 



STRENGTH OF MATERIALS. 2 2 \ 

pxc 



P + %CX A m 
Pi = Corrected breaking load of column. 
C= Crushing strength of material (see Table No. 26). 
A = Metal area of column in square inches. 

Important. — Applying the last formula, the result, P\, 
must always be smaller than P. 

Table No. 27 was calculated by the following formulas : 

( 36000 ) 

Column I. Safe Load = 0.1 X I ti \\ 

I 1 + (w X °- 00025 ) ) 

/ 36000 X 0.07031 ) 
Column II. Safe Load = 0.1X < -, i / Z 2 .. aqqq9c >. [ 

( 36000 ^ 



Column III. Safe Load = 0,1X < . , . Z 2 v A AAA1 . 

/ 36000 X 0.07031 



Column IV. Safe Load = 0.1 X < -. i . L 2 x nfwux 



80000 



Column V. Safe Load = 0,1 X Ji ± ( ^l_ v 00^5 ) 



Column VI. Safe Load = 0.1 X 1 1 _[_ ( ^ 



D 2 
80000 X 0.07031 



p + ( -^ X 0.0025 ) 



( 80000 

Column VII. Safe Load = 0.1X < ri 

U + ijpX 0-0035) 

( 80000 X 0.07031 
Column VIII. Safe Load = 0.1X < . JJ 



1 + ( A- X 0.0035 ) 

V J)'2 



s 



D 1 
5000 



Column IX. Safe Load = 0.1X < , , , LP- v AAA/1 \ 

( 5000 X 0.07031 



Column X. Safe Load = 0.1 X < -, i , Z 2 „ A AA . v 

( X + (^2- X0 - 004 ) 



226 



STRENGTH OF MATERIALS. 



TABLE No. 27.— Safe Load on Pillars Having Square 
Ends Well Fitted. 









(10 is used as Factor ol 


Safety). 










V 

6 

« . 

'•3-S 

•> s 

^ n 
■5 ° 

W) 

c 
<u 


Wrought Iron. 


Hollow 


Cast Iron. 


illar. 


Wood. 
(Spruce or 
white Pine) 


u 
V 

a 

.2 u 

"OT3 

<u — 
V E 

■5 ° 

3 
4) 


Hollow Pillar 


Solid Pillar. 


Pillar. 


Solid I 


O 

s 
»— i 

V 
\- 

u a 

V 3 

P. cr 
s 

3 

o 


u 
l> 

4$ 

u S 

so 

-2 3 

K/ & 

Wc/3 


.3 
o 

a 

1— 1 
l> 

V 3 
P. cr 

*o 

3 
3 
O 


V 

u S 

P.G 
If. v 

go 


o 

s 

V 
u 

1) 3 

P. W 

3 

3 
O 

P-. 


u 

V 
4> 

u g 

&s 
ua <u 

go 
2 g 

3c£ 


O 

3 

HH 
V 

u 
u « 
1) - 

P.& 

73 

3 

3 
O 

Ph 


u 

u 

<u 

* s 

oj-3 

& 3 
in <U 

go 

O § 


3 
(— ( 

<u 

u 

i-i rt 

V 3 

P. cr 

-a 

3 
3 
o 
Ph 


U S 

go 
2 g 

M iS 


A 






















z 


2? 
4 


I. 


II. 


III. 


IV. 


V. 


VI. 


VII. 


VIII. 


IX. 


X. 


z> 
4 


3585 


252 


3567 


252 


7692 


540 


7624 


536 


476 


34 


6 


3567 


251 


3549 


250 


7339 


512 


7105 


500 


438 


30 


6 


8 


3545 


250 


3510 


247 


6896 


484 


6536 


460 


398 


28 


8 


10 


3512 


247 


3462 


243 


6400 


450 


5926 


417 


357 


25 


10 


12 


3475 


244 


3404 


239 


5882 


413 


5314 


374 


317 


22 


12 


14 


3432 


241 


3338 


235 


5369 


371 


4745 


333 


280 


20 


14 


16 


3383 


238 


3266 


230 


4878 


343 


4226 


297 


248 


17 


16 


18 


3332 


234 


3187 


224 


4420 


311 


3749 


264 


218 


15 


18 


20 


3273 


230 


3103 


218 


4000 


281 


3333 


234 


192 


13 


20 


22 


3211 


226 


3018 


212 


3620 


255 


2969 


208 


170 


12 


22 


24 


3147 


221 


2926 


206 


3279 


230 


2653 


186 


151 


10.6 


24 


26 


3080 


216 


2834 


199 


2974 


211 


2376 


167 


135 


9.5 


26 


28 


3010 


211 


2741 


192 


2703 


190 


2137 


150 


121 


8.5 


28 


30 


2938 


206 


2647 


186 


2462 


173 


1928 


136 


109 


7.6 


30 


32 


2866 


201 


2554 


180 


2247 


158 


1745 


123 


98 


6.9 


32 


34 


2793 


196 


2462 


173 


2056 


145 


1585 


111 


88 


6.2 


34 


36 


2719 


191 


2370 


167 


1887 


133 


1440 


102 


81 


5.7 


36 


38 


2645 


186 


2288 


160 


1735 


122 


1321 


93 


74 


5.2 


38 


40 


2571 


181 


2195 


154 


1600 


112 


1212 


85 


67 


4.7 


40 


42 


2498 


176 


2111 


148 


1479 


104 


1113 


78 


62 


4.3 


42 


44 


2426 


171 


2029 


143 


1370 


96 


1028 


74 


57 


4 


44 


46 


2354 


166 


1950 


137 


1272 


89 


952 


67 


53 


3.8 


46 


48 


2284 


161 


1874 


132 


1183 


83 


882 


62 


49 


3.4 


48 


50 


2215 


155 


1800 


127 


1103 


78 


820 


58 


45 


3.2 


50 



This table is intended, in connection with Table No. 24, to 
facilitate calculations for pillars of either wood or iron, and may 
be used with equal advantage for English or metric measures, 
provided both diameter and length are taken by the same system. 

For a round pillar divide the length by the diameter, but 
for a square or rectangular pillar divide the length by the 



STRENGTH OF MATERIALS. 22 7 

smallest side. Find the quotient in the column headed -p~ 
and find the safe load per square unit of area in the cor- 
responding column of the table. Multiply this by the metal 
area of the pillar, and the product is the safe load, with 10 as 
factor of safety, on a pillar well fitted and having square ends. 
For any other kind of ends and any other unit of safety, allow- 
ance must be made as explained on previous pages. 

Example 1. 

Find the safe load in pounds, according to Table No. 27, 
for a round, hollow, cast-iron pillar five feet long, five inches 
outside and four inches inside diameter, having square ends 
well fitted and being evenly loaded. 

Solution: 

Five feet equals 60 inches, and 00 divided by 5 gives 12. 
In the first column, under the heading " Length divided by 
diameter, or smallest side, " is 12, and in that line, in the column 
headed u pounds per square inch " for hollow cast-iron pillars, 
is 5882. The metal area of this pillar is obtained by subtract- 
ing the area of a circle four inches in diameter from the area of 
a circle five inches in diameter (see area of circles, page 196; 
Table, page 209), which is 19.63 —12.57 = 7.06, or practically 
seven square inches, and seven times 5882 equals 41,174 pounds. 

Example 2. 

Find the safe load in kilograms, according to Table No. 27, 
for a round spruce post 2 meters long and 20 centimeters in 
diameter. 

Solution : 

Two meters = 200 centimeters and ^ = 10. The corre- 
sponding constant in the table is 25 kilograms. The area of a 
circle 20 centimeters in diameter is 314.2 square centimeters, 
and 25 times 314.2 = 7855 kilograms, as safe load. 

Example 3. 

What would be the safe load on the same post if it had 
been 20 centimeters square, instead of round ? 

Solution : 

The length is the same ; therefore the length divided by the 
side gives 10, as before, and the corresponding constant is 25 
kilograms, but as the cross-sectional area in square centimeters 
is 20 X 20 = 400, the corresponding load will be 400 X 25 = 
10,000 kilograms as the safe load. 

Note. — It will be noticed that in figuring the strength of 
pillars according to this table, the strength of a square pillar 
will always be to the strength of a round pillar as 1 to 0.7854, 
while theoretically the strength of a square pillar compared to 
that of a round pillar will vary with the length, the extremes 
being 1 to 0.589 for extremely long pillars and 1 to 0.7854 for 



228 STRENGTH OF MATERIALS. 

very short ones. This discrepancy is frequently unimportant in 
practical work, because pillars are usually comparatively short, 
and also because a high factor of safety is always used, but it 
is well to remember and provide for this fact in cases of very 
long pillars. 

Hollow Cast=Iron Pillars. 

By referring to the formulas and considering the laws 
governing the strength of pillars, it is seen that the strength 
of pillars increases very fast by increasing their diameter or 
their sides. In cast-iron pillars this is taken advantage of by 
making them large in diameter and coring out the stock on the 
inside. 

The thickness of the metal may be about T V of the diameter 
of the pillar. In small pillars it must be thicker in order to 
obtain good results when casting. A flange is cast on each end 
to form enough bearing surface, and the pillar is squared off 
very- carefully so that both ends are square with the center-line. 
This is an important point, as the strength is enormously 
destroyed by squaring the ends carelessly and thereby bringing 
the load to act corner-ways on the pillar. 

Table No. 28 was calculated by the formula: 

r 80000 X metal area \ 
Safe Load = 0.1 X j x + ( ^ x 0.0025) j 

and the result obtained reduced to long tons (2240 pounds). 
Ten is thus used as a factor of safety : both ends of the pillar 
are supposed to be square and evenly loaded. For other shapes 
of ends, mode of loading, or other factors of safety, propor- 
tional allowance must be made. For instance, if 15 is required 
as factor of safety, allow only two-thirds of the load given in 
the table. 

If the pillar has only one square end and one round end, 
allow only two-thirds as much load. If it has both ends 
rounded, or. which is the same, if the ends have only a very 
imperfect oearkig. allow only one-third as much load. 

Weight of Cast=Iron Pillars. 

The weight of a cast-iron pillar may be calculated by the 
formula: 

W=- (D 2 — d 2 ) X L X 2.45 
IV = Weight of pillar in pounds. 
D = Outside diameter in inches. 
d= Inside diameter in inches. 
L = Length of pillar in feet. 

The weight given in Table No. 28 was calculated by this 
formula, and the length taken as one foot. 



STRENGTH OF MATERIALS. 



220 



TABLE No. 28.— Safe Load on Round Cast=Iron Pillars. 



£3 V 


U) i 

1> ^ 








Length of Pillars in Feet. 


^ 


t/1 <-> 

1) .-. 


< -5 


'"" 0-.£j= 
































«"" 


C C 


— C c 


-<=-rJ tr Wi 






^j 


w 


+j 


^j 


^j 


4J 


jj 




C .S 


^ ■" 


rt— ".i3 


MpgC 






u 


<U 


V 


V 


u 


V 


u 


4> 






■h b ,0 «i 


V 


D 


u 


W 


u 


V 


V 


V 


V 


u 


0) 1. 






41 3 fa J 


fa 


fa 


fa 




fa 


fa 


fa 

CO 


fa 

00 


fa 



fa 

CM 


fa 

>* 


W v 






CO 


00 












<N 


<M 


<M 


4 


% 


5.49 


17.14 


11 


8.1 


6.1 


4 


X 


7.56 


23.90 


15.2 


11.3 


8.5 
















5 


% 


7.07 


22.06 


16.8 


13.3 


10.4 


8.3 














5 


% 


10.01 


31.23 


24 


19 


15 


12 














6 


% 


8.64 


26.95 


23 


19 


15.5 


12.6 


10.4 












6 


X 


12.37 


38.59 


33 


27 


22 


18 


15 












6 


n 


14.09 


43.96 


37 


31 


25 


21 


17 












6 


1 


15.71 


49.01 


42 


35 


28 


23 


19 












6 


ly* 


17.23 


53.76 


47 


40 


32 


26 


22 












7 


y& 


12.52 


39.06 


36 


31 


26' 


22 


19 


16 










7 


x 


14.73 


45.96 


42 


36 


31 


26 


22 


19 










7 


y* 


16.84 


52.54 


48 


41 


35 


29 


25 


21 










7 


1 


18.85 


58.90 


54 


46 


39 


33 


29 


24 










7 


ty s 


20.76 


64.77 


60 


52 


44 


37 


32 


27 










8 


U 


17-08 


53.29 


51 


45 


39 


34 


29 


25 


22 








8 


n 


19.59 


61.12 


59 


52 


45 


39 


34 


29 


25 








8 


1 


21.99 


68.64 


66 


58 


51 


44 


38 


33 


28 








8 


iy% 


24.30 


75.82 


73 


64 


56 


48 


42 


36 


31 








8 


IX 


26.51 


82.71 


79 


70 


61 


52 


45 


39 


34 








8 


1/8 


28.62 


89.29 


86 


76 


66 


57 


48 


42 


37 








9 


x 


19.44 


60.65 


60 


54 


49 


43 


37 


33 


29 


24 






9 


H 


22.33 


69.67 


69 


63 


56 


49 


43 


38 


33 


29 






9 


1 


22.13 


78.40 


78 


71 


63 


55 


48 


42 


37 


33 






9 


1>8 


27.83 


86.83 


87 


78 


69 


62 


53 


47 


41 


36 






9 


IX 


30.43 


94.94 


95 


85 


76 


67 


58 


51 


45 


39 






9 


1^8 


32.94 


102.77 


102 


92 


82 


72 


63 


55 


48 


43 






9 


1^ 


35.34 


110.26 


110 


99 


88 


78 


68 


59 


52 


46 






9 


IX 


39.86 


124.36 


126 


113 


100 


90 


78 


67 


60 


51 






10 


n 


25.09 


78.28 


80 


73 


67 


60 


53 


47 


42 


37 


34 




10 


1 


28.28 


88.23 


90 


83 


75 


67 


60 


53 


47 


42 


38 




10 


1^ 


31.37 


97.87 


100 


92 


83 


74 


66 


58 


52 


47 


42 




10 


IX 


34.37 


107.23 


110 


101 


91 


82 


73 


64 


57 


51 


47 




10 


1^8 


37.26 


116.25 


119 


109 


98 


88 


79 


69 


62 


55 


51 




10 


1/2 


40.06 


124.99 


128 


117 


106 


95 


85 


75 


67 


59 


54 




10 


IX 


45.36 


141.52 


146 


133 


122 


109 


97 


85 


77 


67 


60 




11 


1 


31.42 


98.04 


102 


95 


87 


79 


71 


64 


58 


52 


48 


43 


11 


1# 


34.90 


108.89 


114 


105 


96 


88 


79 


71 


64 


58 


53 


48 


11 


IX 


38.29 


119.46 


125 


116 


106 


97 


87 


78 


70 


63 


58 


52 


11 


1^41.58 


129.73 


135 


126 


115 


105 


94 


85 


76 


68 


62 


56 


11 


1^44.77 


139.68 


146 


136 


124 


113 


102 


92 


82 


74 


68 


61 


11 


IX 50.86 


158.68 


166 


156 


142 


129 


178 


106 


94 


86 


79 


71 


11 


2 |56.55 


176.44 


186 


176 


160 


147 


134 


120 


106 


98 


90 


81 



230 



STRENGTH OF MATERIALS. 



TABLE No. 28. — (Continued). 



5 v 


to 


<« 


c u 




Length of Pillars in Feet. 


^8 

12 


tfi c 

c C 
^.- 

.3*3 

1 


'- 0) 
< "G 

t> ^. 


-G-a *■ M 
^ = fa J 






4> 
V 

fa 


I) 

fa- 
ce 


V 

V 

fa 




V 

u 

fa 


V 

V 

fa 


1) 

fa 


t3 

fa 

00 


fa 

O 


V 

V 

fa 


u 

fa 

* 


34.46 107.51 


115 


107 


99 


92 


83 


76 


68 


62 


58 53 


12 


1M 


38.34 


119.62 


128 


119 


108 


102 


92 


84 


78 


69 


63! 58 


12 


IX 


42.12 


131.41 


141 


131 


119 


112 


101 


93 


84 


76 


70 64 


12 


ix 


45.80 


142.90 


153 


142 


129 


121 


110 


101 


91 


82 


75 69 


12 


1% 


49.39 


154.10 


165 


154 


139 


131 


119 


109 


99 


89 


82 75 


12 


iy A 


56.26 


175.53 


189 


178 


159 


150 


137 


125 


115 


103 


94 


85 


12 


2 


62.74 


195.75 


213 


201 


179 


170 


155 


141 


131 


117 


106 


96 


13 


1 


37.97 


117.53 


127 


119 


111 


104 


97 


87 


79 


73 


67 


61 


13 


IX 


41.94 


130.85 


142 


134 


126 


117 


109 


101 


91 


82 


75 


69 


13 


IX 


46.11 


143.86 


158 


149 


140 


130 


121 


112 


101 


91 


84 


77 


13 


IK 


50.19 


156.59 


174 


163 


154 


144 


133 


122 


111 


101 


93 


84 


13 


IK 


54.16 


168.98 


190 


178 


168 


157 


145 


133 


121 


110 


101 


92 


13 


IK 


61.82 


192.88 


214 


201 


189 


176 


164 


151 


137 


124 


114 


104 


13 


2 


69.09 


215.56 


227 


224 


210 


195 


182 


168 


152 


137 


126 


116 


14 


1 


40.94 


127.60 


138 


131 


123 


115 


109 


101 


92 


85 


78 


72 


14 


IK 


45.50 


141.96 


153 


145 


139 


130 


121 


112 


103 


94 


87 


80 


14 


IX 


50.07 


156.31 


168 


160 


153 


143 


133 


123 


113 


104 


95 


88 


14 


IK 


54.54 


170.04 


183 


174 


167 


156 


145 


134 


123 


113 


104 


95 


14 


iy 2 


58.78 


183.67 


198 


189 


180 


168 


156 


145 


133 


122 


112 


103 


14 


IX 


67.31 


210.00 


226 


216 


206 


192 


179 


166 


152 


140 


128 


118 


14 


2 


75.36 


235.12 


254 


242 


232 


216 


201 


186 


171 


157 


143 


133 


15 


1 


43.99 


137.28 


150 


143 


136 


128 


120, 


112 


104 


96 


90 


82 


15 


ik 


49.04 


153.19 


167 


159 


152 


143 


136 


125 


116 


108 


100 


93 


15 


IX 


54.00 


168.48 


184 


175 


167 


157 


148 


138 


128 


119 


110 


102 


15 


ih 


58.85 


183.77 


201 


191 


182 


172 


161 


151 


140 


130 


120 


111 


15 


ik 


63.62 


198.74 


217 


207 


197 


186 


175 


163 1 


151 


141 


130 


120 


15 


IX 


72.85 


227.45 


284 


236 


225 


212 


202 


190^ 


175! 


160 


148 


137 


15 


2 


81.68 


254.81 


279 


265 


253 


237 


229 


216 


197 


179 


166 


154 


16 


1 


47.12 


146.95 


160 


156 


150 


140 


133 


125 


117: 


110 


101 


93 


16 


IK 


52.57 


164.11 


179 


173 


167 


158 


148; 


139, 


130 


122 


116 106 


16 


IX 


57.92 


180.65 


198 


191 


185 


174 


163 


153 


143 


135 


128 


117 


16 


IK 


63.18 


197.18 


216 


209 


202 


190 


178 


167 


152 


147 


139 


128 


16 


IK 


68.33 


213.10 


234 


227 


220 


206 


193 


181 


170 


160 


151 


138 


16 


IX 


78.34 


244.29 


26S 


258 


250 


237 


224 1 


207, 


199, 


183 


173 


158 


16 


2 


87.96 


274.56 


301 


290 


281 


267 


255, 


233 


218 


206 


194 


177 



The length of cast-iron pillars, as a rule, ought not to ex- 
ceed 20 to 25 times their diameter. Cast-iron pillars, when 
heavily loaded, are apt to be broken if struck by a blow 
sidewise. 



STRENGTH OF MATERIALS. 



231 



Wrought Iron Pillars. 

In important work, cast-iron pillars are rapidly going out of 
use. Wrought iron pillars are now made which compare 
favorably in price and are far more reliable than those of cast 
iron. For full information regarding weight and strength of 
wrought iron pillars and Z bar columns, see manufacturers' 
catalogues. 

Wooden Posts. 

Table No. 29 was calculated by the formula : 
( 5000 X Area ) 

Safe load = 0.1 



X 0.004 



) 



and the result divided by 2240. 

L = Length of post in inches. 

D = Side of post in inches. 
TABLE No. 20. — Safe Load in Tons on Square Pine or 
Spruce Posts Having Square Ends Well Fitted. 

(10 as Factor of Safety). 



at: 

c ■*■< 

-1.2 


Side of Post 


in Inches. 


1 in. 


2 in. 


3 in. 


4 in. 


5 in. 


6 in. 


7 in. 


8 in. 


9 in. 


10 in. 


12 in. 


1 


0.14 


0.78 


1.89 


3.45 


5.00 


8.00 


10.00 


12.50 


16.00 


20.00 


30.00 


2 


0.06 


0.56 


1.59 


3.12 


5.00 


8.00 


10.00 


12.50 


16.00 


20.00 


30.00 


3 


0.03 


0.39 


1.27 


2.70 


4.62 


7.00 


10.00 


12.50 


16.00 


20.00 


30.00 


4 


. . 


0.32 


0.99 


2.27 


4.08 


6.38 


9.21 


12.50 


16.00 


20.00 


30.00 







0.19 


0.77 


1.8S 


3.54 


5.74 


8.45 


11.21 


15.35 


19.51 


29.22 


6 


. 


0.14 


0.60 


1.56 


3.05 


5.00 


7.68 


10.80 


14.35 


18.44 


28.10 


7 


. . 




0.49 


1.29 


2.62 


4.50 


6.94 


9.91 


13.41 


17.40 


26.87 


8 


. . 


m # 


0.39 


1.08 


2.25 


4.00 


6.24 


9.08 


12.44 


16.32 


25.50 


10 


. . 


. m 


0.27 


0.78 


1.69 


3.09 


5.03 


7.53 


10.58 


14.18 


22.95 


12 










1.29 


2.43 


4.06 


6.25 


8.96 


12.22 


20.40 


14 








0.44 


1.00 


1.94 


3.31 


5.19 


7.57 


10.50 


18.02 


16 






. # 


. . 


0.90 


1.58 


2.73 


4.34 


6.43 


9.04 


15.88 


18 










0.80 


1.30 


2.27 


3.66 


5.49 


7.80 


14.00 


20 












1.09 


1.92 


3.11 


4.72 


6.76 


12.36 


22 












0.90 


1.63 


2.66 


4.17 


5.89 


10.95 


24 


















3.55 


5.17 


9.73 


26 


. . 


# # 


a # 


# m 


. . 


. . 


1.22 


2.01 


3.11 


4.56 


8.6S 


28 


. . 


. , 


# , 


. 


. . 


. . 




1.78 


2.68 


4.05 


7.77 


30 
















1.56 


2.38 


3.56 


6.99 


32 


















2 18 


3.23 


6.30 


34 


















1 96 


2.91 


5.71 


36 




















2.64 


5.18 


3S 




















2 39 


4.74 


40 






















4.35 



2 $2 STRENGTH OF MATERIALS. 

The preceding Table gives the safe load in long tons corre- 
sponding to a square post of the dimensions of sides given at 
top of the columns, and lengths given in the first column. For 
round posts the load should be 0.75 to 0.6 of the given load de- 
pending upon the length of post. 

Example. 

What size of post is required, with 10 as factor of safety, 
to support a load of five tons, when the length of the post is 16 
feet? 

Solution : 

In the column headed "Length of post in feet" find 16, 
and in line with 16 find the numbers nearest to five tons, which 
are 4.34 and 6.43. Thus, a post 16 feet long and 8 inches square 
will support 4.34 tons, and a post 16 feet long and 9 inches 
square will support 6.43 tons. It is, therefore, best to select a 
post 9 inches square. 

To Calculate the Strength of Rectangular Posts from 

the Table. 

Find, in the Table, the strength of the post according to 
its smallest side, and increase the tabular value in proportion 
to the largest side of the post. 

Example. 

What is the strength, with 10 as factor of safety, of a spruce 
post 10 feet long, 6 inches thick, 8)4 inches wide, with square 
ends well fitted, calculated by Table No. 29. 

Solution : 

In the Table we find the strength of a post 10 feet long 
and 6 inches square to be 3.09 tons. Therefore, when the pillar 
is 6 inches thick and 8% inches wide its corresponding strength 

will be 3.09 X ^ = 4.38 tons. 

It is a waste of material to use a post of rectangular cross- 
section. For example, this post is 6 X 8% inches = 51 square 
inches of cross-section and will support 4.38 tons, but a post of 
the same length and 7X7 inches = 49 square inches of cross- 
section, will support 5.03 tons. ( See Table No. 29). 



To Obtain the Weight of Pillars in Kilograms per Meter 
when the Weight in Pounds per Foot is Known. 

Multiply the weight in pounds per foot by the constant 
1.4S82, and the product is the weight in kilograms per meter. 



STRENGTH OF MATERIALS. 233 



TRANSVERSE STRENGTH. 

A beam placed in a horizontal position, fastened at one 
end and loaded at the other, is exposed to transverse stress, 
and will usually bend more or less, as shown (exaggerated) in 

Fig. 1, before it will break. The line 
F|G - *• a b is called the neutral line, and all 

^____^ ., ^ fibres above the neutral line are exposed 

""!::i^^5^rrZIIl to tensile stress, and all fibers below 

" i ^~ v ON/\ are exposed to crushing stress, but the 

N/\) neutral fiber is neither stretched nor 

/^k compressed. A line drawn in a hori- 

/ \ zontal direction, at right angles to, and 

* ^ through the neutral line, is called the 

neutral axis with reference to this particular place of the section 
of the beam. The neutral axis is considered to pass through 
the center of gravity of the section, which, for beams of round, 
square or rectangular section, is always in the geometrical cen- 
ter. Therefore, all beams of such section will have an equal 
amount of material on the upper and under side of the neutral 
axis, but it is not always desirable for all materials or for all 
kinds of load to have an equal amount of material on both the 
side exposed to compression and that exposed to tension. For 
instance, cast-iron beams are usually made in T formed sec- 
tion and should always be laid so that the largest web is 
exposed to tensile stress, because cast-iron offers much 
more resistance to compression than it does to tension. Cast- 
iron beams of such section ought, therefore, to be laid in this 
position (T), if fastened at one end and loaded at the other, 
but should be laid in this position ( X ), if they are supported 
under both ends and loaded between the supports. If 
this is taken into consideration in placing a cast-iron beam, its 
ultimate transverse breaking strength is greatly increased, but 
under a moderate load the deflection will be practically equal 
in either position, because as long as the load is small, well 
within the elastic limit, cast-iron will stretch under tensile stress 
as much as it will compress under an equal amount of crushing 
stress; therefore, the modulus of elasticity for tension and 
compression of cast-iron is considered to be equal, but under 
increased crushing load the compression becomes less in pro- 
portion to the load until the point is reached when the cast-iron 
can not compress more, and the casting will break. The 
ultimate crushing strength of cast-iron is five to six times as 
much as its ultimate tensile strength. 

A beam supported under both ends and loaded in the 
middle will carry four times as great a load as another beam of 
the same size and material fixed at one end and loaded at the 



234 



STRENGTH OF MATERIALS. 





1 foot. 




1 


25 lbs. 


FIG. 2. 




I n .. ■. 


A r 


\ 


&4^ 


\m&> J 



100 lbs. 



other. This may be understood by 
referring to Fig. 2, as when the beam 
is one foot long and loaded with 
100 pounds in the middle, each half 
of the beam supports only 50 pounds, 
and this 50 pounds acts only upon an 
arm y 2 foot long, consequently it 
exerts no more force toward break- 
ing this beam than the 25 pounds 
would upon the end of the other 
beam one foot long. 

A beam twice as wide as another and of the same length, 
thickness, and material, will carry twice the load, because the 
wide beam could, of course, be split into two equal beams ; 
consequently it must, as a whole 
beam, have twice the strength of 
another one of the same material 
but of only half the width. 

50 lbs 



1^1 it. 



.2 feet. 



100 lbs. 



Fig. 3. 



A beam twice as long as another 
will break under half the load. This 
is seen by referring to Fig. 3, be- 
cause 50 pounds on an arm two feet 
long will balance 100 pounds on an arm one foot long. 

A beam twice as thick as another, of the same material, 
length and width, will carry four times the load. ( See Fig. 4). 

Suppose the weight a is act- 
ing on the arm 6, tending to swing 
it around the center c, and this 
action being counteracted by the 
weights g and k, also by the 
arrows e and f. If the weight h 
is taking hold twice as far from 
the center as the weighty, it will 
offer twice the resistance against 
swinging the beam that g will; 
and exactly the same with the 
arrows/" and e. Consider the line c b as the neutral fiber, the 
arrows e and f as representing the fibers resisting crushing, 
and the weights £• and h as representing the fibers resisting tensile 
stress. It will be understood that if the fibers are twice as far 
above or below the neutral fiber they are in a position to orfer 
twice the resistance to the breaking action of the load ; but a 
beam of twice the thickness has not only its average fiber twice 
as far from the neutral point, but it has also twice the area or 
twice as many fibers, consequently the result must be that it 
can resist four times the load. 




Oft 



STRENGTH OF MATERIALS. 



235 



For instance : The beam a in Fig- 
ure 5 is four times as strong as the 
beam b, if placed on the edge, as shown 
in the figure, and loaded on the top ; 
but a would be only twice as strong as 
b if it was laid on the side and loaded 
on top. 



U-4in.»( 





Formulas and Rules for Calculating Transverse Strength 

of Beams. 

The fundamental formula for transverse stress in beams is: 
Bending Moment = Resisting Moment. 



The bending moment for a beam fixed at one end and 
loaded at the other (see Fig. 1) is obtained by multiplying the 
load by the horizontal distance from the neutral axis to the 
point where the load is applied. The distance is taken in 
inches and the load in pounds. 

The resisting moment is obtained by multiplying the mo- 
ment of inertia by the unit stress, tensile or compressive, upon 
the fiber most remote from the neutral axis, and dividing the 
product by the distance from this fiber to the neutral axis. 

The theoretical formula for the transverse strength of a 
beam fastened in a horizontal position at one end and loaded at 
the extremity of the other end, as shown in Fig. 6, is, 

SX I 



P 



L X a 



When the beam is fastened at one end and loaded evenly 
throughout its whole length, as shown in Fig. 7, the formula 
will be, 

P = 2 X S X [ 
L X a 

When a beam is placed in a horizontal position and sup- 
ported under both ends and loaded in the middle (see Fig. 8) 
the formula is, 

S X r 



P = ±X 



L X a 



When a beam is placed in a horizontal position and sup- 
ported under both ends and loaded throughout its whole length 
(see Fig. 9), the formula will be, 

L Xa 



236 STRENGTH OF MATERIALS. 

When a beam is laid in a horizontal position, fixed at both 
ends and loaded in the middle between fastenings (see Fig. 10), 
the formula will be, 



L X a 

When a beam is laid in a horizontal position, fixed at both 
ends and the load evenly distributed over its whole length (see 
Fig. 11), the formula will be, 

P = 12 X S X l 

L X a 

P = Breaking load in pounds. 

S= Modulus of rupture, which is 72 times the weight, in 
pounds, which will break a beam one inch square and one foot 
long when fixed in a horizontal position, as shown in Fig. 6, 
and loaded at the extreme end, and which may be taken as 
follows : 

Cast-Iron. 36.000. 

Wrought Iron. 50.000. 

Spruce and Pine, 9,000. 

Pitch Pine, 10,000. 

These are the nearest values, in round numbers, of 72 times 
the average value of the constant given in Table No. 30. 

For the safe load, S may be taken as follows : 
For timber. 1,000 to 1,200 pounds. 
For cast-iron, 3.000 to 5,000 pounds. 
For wrought iron, 10,000 to 12.000 pounds. 
For steel, 12.000 to 20,000 pounds. 

L = Length of beam in inches. 

a ■==■ The distance in inches from the neutral surface of 
the section to the most strained fiber. 

/=: Rectangular moment of inertia. 

The tables on pages 237 and 238 give the moment of inertia 
about the neutral axis XV. and the distance a, for a few of the 
most common sections : 

(For explanation of moment of inertia and center of gravity 
see page 293). 

These formulas have the great advantage of being theoretic- 
ally correct for beams of any shape of cross-section, made from 
any material, providing the load is within the elastic limit of 
the beam, and a correct constant is used for 6* and the correct 
value obtained for the moment of inertia, 



STRENGTH OF MATERIALS. 



237 




a — 



HI 
6 



a = 



I_ B x (H*—h*) 
a ~ Qx H 




a ==■ 



/ __ H 2 x B 
a 6 





I — 



/== 



H± 



Bxff* 

36 



12 

a—\Hx VT 
I 



a 



= 0.118 H* 



a=y 3 H\ av=YiH 
I _ B X H 2 

ax 12 

/ BXH 1 




a = 



/ 



24 



— = 0.0982 x D 3 
a 




a =- 



— = 0.0982 D x D' 2 
a 




\—*-\ 




64 

2 
£_{DxD^ — dxd^ 
a~ 32 Z? 



h d >] 



64 

2 

/ _ (Z> 4 — </*) 7T 

<z 32 £> 



2 3 8 



STRENGTH OF MATERIALS. 




riiF-r 



x -f-T 

8 



11 



wzz* 



a = 



H 




QH 




H~T 



/ = 



12 





_/_ . B H* + bh* 

a QH 







< B- 



T 1 



KZTT! A 1 



PTTT 



r 






T i 



#i 



_B H 1 — 2 Hbh + bh 2 
>2 (B H — b h) 

_ BH 2 — bh 2 
2{B H—bh) 



I — 


AB H* — bh*)* — 4.B H b h (H—hf 




12 (B H — bh) 


I _ 


(B H 2 — b h?f — ±BHbh(H — hy 2 


a\ 


6 {B H 2 — 2 H b h + b h?) 


I _ 


_ {B H* — b h 2 ) 2 — A B H b h (H — Kf 



Q(B H^ — b If-) 



STRENGTH OF MATERIALS. 



2 39 



Beams of symmetrical section, as square, round, elliptical, 
or H section, may be calculated on theoretically correct prin- 
ciples in a simpler way, obviating the use of the moment of 
inertia and the modulus of rupture, as explained below. 

For a beam fixed at one end and loaded at the other, 
CXH 2 XB „ fig. e. 



P = 



L 



L 
CX H 2 X B 



T= 



-L 




-^1 



H=JPXL 



B 



CX B 

PXL 
CX H 2 



C = 



H 2 X B 



r x l 



When beam is round, 
Diameter = 3 -* / 



PXL 

C X 0.589 



When beam is square, 
Side = 3 J ^XZ 

P = Breaking load in pounds. 

H ' =. Thickness or height of beam in inches. 

B = Width of beam in inches. 

L = Length of beam in feet. 

C= Constant which is obtained from experiments, and is 
the weight in pounds which will break a beam 1 foot long and 
1 inch square fixed at one end and loaded at the other. Con- 
stant C is given in Table No. 30. 

A rectangular beam fixed at one end 
and loaded evenly throughout its whole 
length will carry twice the load of a beam 
fixed at one end and loaded at the other; 
therefore, 



Fig. 7 



P = 



2 X C X H 2 X B 



L 




For a rectangular beam supported under both ends and 
loaded at the center, fig. 8. 



P = 



4 X C X H 2 X B 
L 




A rectangular beam supported under both ends and loaded 
evenly throughout its whole length will carry twice the load of 



240 



STRENGTH OF MATERIALS. 



a beam supported under both ends and loaded at the center; 
therefore, FlG g 



P 



8 X C X If 2 X B 
L 



For a beam fixed at both ends 
and loaded at the center, 



P = 



8 X C X 7/ 2 X B 
L 



For a beam fixed at both ends 
and the load distributed evenly 
throughout its whole length, 



P — 



12 X C X H* X B 
L 




Each letter in these formulas has the same meaning as in 
formula for Fig. 6, page 239, and each formula may be transposed 
the same as that formula. The most convenient way is, in each 
case, to multiply the numerical value of Cfrom Table No. 30, by 
its proper coefficient according to mode of loading, before it is 
inserted in the formula. 

Note. — A square beam laid in this ■ position has 40 % 
more transverse strength than the same beam laid in 
this ♦ position. 

TABLE No. 30. — Constant C. 

Giving the weight in pounds which will break a beam one 
foot long and one inch square which is fastened at one end, in a 
horizontal position, and loaded at the other end. 



Material. 


Very Good. 


Medium. 


Poor. 


Wrought iron,* 
Cast-iron, 
Spruce and Pine, 
Pitch pine, 
Granite, 


750 
650 
160 
225 


600 
500 
125 
150 
25 


500 

400 

90 

100 



* A wrought iron beam or bar will not actually break under these conditions, 
but, as it will bend so much that it becomes useless, it is considered to be equivalent 
to the breaking point, 



STRENGTH OF MATERIALS. 



24I 



The following formulas will apply to the strength of beams 
of the shape shown in the adjacent sectional cuts. These for- 
mulas pertain only to the ultimate breaking strength of beams, 
and have nothing to do with deflection, which follows entirely 
different laws. 



SOLID SQUARE BEAMS. 



P — 

L — 



C X If 3 


L 


CXH* 



Fig. 12. 



C — 



p 




PX L 



=v 



/f=lPXL 



HOLLOW SQUARE BEAMS. 



C X (H± — h 4 ) 
LXH 



Fig. 13. 



in? 



p= 



L — 



SOLID RECTANGULAR BEAMS. 

CX B X H 2 fig. 14 

7 



L 



CX B X H 2 
P 



h^^p^L 




V 



CXH 2 



c 



B "I 



p = 



C X B 

HOLLOW RECTANGULAR BEAMS. 

CX {B X H* — bXh?) 
HXL 



L— CXjBXH* —b X h*) 
PX H 



P XL 
B X H 2 






II 



-B- 
FlG. 15. 



P = 



z = 



SOLID ROUND BEAMS. 

0.589 C X D* fig. 16. 



L 



0.589 CX D* 
P 




D = 



-V-£ 



XL 



589 C 



242 



STRENGTH OF MATERIALS. 




Fig. 18. 




HOLLOW ROUND BEAMS. 

P — 0-589 CX(Z^ 4 -^) 
L X D 



SOLID ELLIPTICAL OR OVAL BEAMS. 

p _ 0.589 C X DxXD 2 
L 




Fig. 20. 



¥fF] 



— B— 



HOLLOW ELLIPTICAL OR OVAL BEAMS. 



p _ 0.589 C(Di X D 3 — d x X d s ) 
LD 



I BEAMS. 

As a general rule, wrought iron X beams 
should always be selected of such size that their 
depth is not less than one-twenty-fourth of the 
span; and their strength may be calculated by the 
formula : 

p __ CX (B X H* —2b X h?) 
LXH 



In the preceding formulas : 

P '= Breaking load when beam is fastened at one end and 
loaded at the other. 

L = Length of beam in feet. 

C = Constant, and is the load in pounds which will break a 
bar one inch square and one foot long when fastened at one 
end and loaded at the other, and may be obtained from 
Table No. 30. 

These formulas give the breaking load when the beam is 
fastened at one end and loaded at the other, but for other 
fastenings and loads C must be multiplied by either 2, 4, 6, 8, 
or 12, depending upon conditions. (See pages 239 and 240). 



Strength of materials. 



243 



Fig 21 



To Find the Transverse Strength of Beams when Their 

Section is Not Uniform Throughout the 

Whole Length. 

Example. 

A beam made of wrought iron is fastened at one end and 
loaded at the other (as shown in Fig. 21). The largest part 
is 5 inches in diameter and the smallest part is 4 inches in 
diameter. Where will it break? 
and what is the breaking load ? 

Note. — Naturally, the beam 
will break at either A ox B\ 
therefore, calculate first the break- 
ing load of a round beam of 
wrought iron 4^ feet long and 5 
inches in diameter, next a beam 3 
3 feet long (the distance from P to 
B). and 4 inches in diameter. 

Solving for strength at A : 

p — £> s 0.6 C 




P = 



L 
5 3 X 0.6 X 600 

±% 
P = 10,000 pounds. 

Solving for strength at B : 
4 3 X 0.6 X 600 



P — 



P = 



64 X 360 



P = 7680 pounds. 

Thus, the weakest point of the beam is at B, where its cal- 
culated breaking load is only 7860 pounds, while the calculated 
breaking load at A is 10,000 pounds. 

When a beam is not of uniform section throughout its 
whole length and is supported under both ends and loaded 
somewhere between the supports, calculate first the reaction 
on each support ; then consider the beam as if it was fastened 
by the load, and consider the reaction at each support as a load 
at the free end of a beam of length and section equal to the 
length and section between its load and support. 

Example. 

The largest diameter of a round cast-iron shaft is 3 inches 
and the smallest diameter is 2 inches. The length, mode of 



*44 



STRENGTH OF MATERIALS. 



loading and support is as shown in Fig. 22. Where will it 
break ? and what is the breaking: load ? 



Fig. 22. 




Solution : 

The reaction at y will be Y% of the load and the reaction 
at x will be % of the load. The beam will evidently break 
either at a. b or n. (Find constant C in Table No. 30 and 
multiply by 0.6. because the beam is round). 

Solving for strength at n : 
2 3 X 500 X 0.6 



n p = 



p = 



3 
S X 300 X 8 



3X3 
P = 2133 >£ pounds. 

Solving for strength at b : 
-, „ 3 3 X 500 X 0.6 

78 P = z 



P = 



27 X 300 X 8 
5 X 3 

P = 4320 pounds. 



Solving for strength at a : 
2 3 X 500 X 0.6 



H P 



8 X 300 X 8 



P — 

5 X 2 

P = 1920 pounds. 

Thus, the weakest place in the beam is at a. where it will 
break when loaded at b with 1920 pounds. If the load is moved 
nearer u. it will at a certain point exert the same breaking stress 
on both a and n. 

Regular beams of this kind are seldom dealt with, but shafts 
or spindles of similar shape and loaded in a similar manner are 
frequently used, and their strength and stiffness may be calcu- 
lated and their weakest spot ascertained by this way of reason- 
ing, which applies as well to hollow as to solid shafts and 
spindles made from wrought iron, steel or cast-iron. 




STRENGTH OF MATERIALS. 245 

Beams fastened at one end and loaded at the other may be 
reduced in size toward the loaded end and still have the same 
strength. Suppose the beam to be fastened in the wall at X 
(Fig. 23) and loaded at the other end with a given load, this load 
will then have the greatest breaking effect upon 
the beam at X\ at half way between X and d fig. 23. 

the load has only half the breaking effect, at c 
only one-quarter the effect. Therefore, the beam 
may be tapered off toward b in such proportion 
that the square of the height a is equal to 
three-quarters the square of the height at X. 
The square of the thickness at b is one-half the 
square of the thickness at X, and the square of the thickness at 
c is one-quarter the square of the height at X. 

Example. 

An iron bracket is four feet long, projecting from a wall 
(as Fig. 23). It is strong enough when 24 inches high at X. 
How high will it have to be at a, b and c ? 

Solution : 

X — 24 2 = 576 Height at X = 24" 

a = V % X 576 = V432 Height at a — 20.78" 
b — V / 2 X 576 = V288 Height at b = 16.97" 
c — V % X 576 = Vl44 Height at c = 12" 

The curved boundary line of such a beam is a parabolic 
curve, because the property of a parabola is that the square of 
the length of any one of the vertical lines (ordinates) is in pro- 
portion as their distance from the extreme point d. By this 
construction one-third of the material may be 
saved and the same strength be maintained. FlG - 24 - 

If the load is distributed along the 
whole length of the bracket instead of at its 
extreme end, it would have the form shown in 
Fig. 24. 

Square and Rectangular Wooden Beams. 

The strength increases directly as the width and as the 
square of the thickness. The strength decreases in the same 
proportion as the length of the span increases. 

Example 1. 

Find the ultimate breaking load in pounds of a spruce 
beam 6 inches square and 8 feet long, when supported under 
both ends and loaded at the center. 




246 STRENGTH OF MATERIALS. 

Solution : 

4 CX H s 
L 
p_ 4X125X6X6X6 



P = 13,500 pounds. 

Note.— C is obtained from Table No. 30, and is multiplied 
by 4 because the beam is supported under both ends and loaded 
at the center. The beam is square ; therefore the cube of the 
thickness is equal to the square of the thickness multiplied by 
the width. Consequently, for a square beam (thickness) 3 or 
(width) 3 or square of thickness multiplied by width is the same 
thing. 

Example 2. 

Find the load which will break a spruce beam 8 inches 
thick, 4 l / 2 inches wide, and 8 feet long, supported under both 
ends and loaded at the center. 

Solution : 

4 CX B X H 2 



P — 



P = 



L 

4 X 125 X 4]/ 2 X 8 X 8 



P = 18,000 pounds. 

Example 3. 

Find the load which will break the beam mentioned in 
Example 2, if beam is laid flatwise. 

4X125X8X4^X4^ 

8 

P = 10,125 pounds. 

In the first example the beam is square, 6" X 6" = 36 
square inches, and its calculated breaking load is 13,500 pounds. 
In the second example the beam is rectangular, 8" X 4 l / 2 " = 
36 square inches, and laid edgewise its figured breaking load 
is 18,000 pounds. In the third example the same beam is laid 
flatwise, and its breaking load is only 10,125 pounds. Thus, by 
making a beam deep it is possible to secure great strength with 
only a small quantity of material, but the limit is soon reached 
where it will not be practical to increase the depth at the ex- 
pense of the width, because the beam will deflect sidewise and 
twist and break if it is not prevented by suitable means. The 



STRENGTH OF MATERIALS. 247 

strongest beam which can be cut from a round log is one hav- 
ing the thickness 1% times the width. The stiffest beam cut 
from a round log has its thickness l T 7 o times its width. The 
best beam for most practical purposes which can be cut from a 
round log has its thickness \y 2 times its width; for instance, 
4 X 6, or 6 X 9, or 8 X 12, etc. The largest side in a beam 
having its thickness \y 2 times its width which can be cut from 
a round log is found by multiplying the diameter by 0.832. The 
diameter required in a round log to be large enough for such a 
beam is found by multiplying the largest side of the beam by 
1.2; for instance, the diameter of a round log to cut 6" X 9" will 
be 9" X 1.2 = 10.8 inches, or the diameter of a round log re- 
quired to cut 8" X 12" will be 12" X 1.2 = 14.4 inches, etc. 



To Calculate the Size of Beam to Carry a Given Load. 

Most frequently the load and the length of span are known 
and the required size of beam is to be calculated. For a rect- 
angular beam there would then be two unknown quantities, 
the width and the thickness, but if it is decided to use a beam 
having its thickness 1 y 2 times its width, the thickness may be 
expressed in terms of the width. 

H= Thickness. 

B = Width. 

H=l% B 

Use formula for rectangular beams, page 239, and it will 
read, 

p= CX (iy 2 BfXB 
L 

This will reduce to, 

C X2% X B 3 



P = 

This will transpose to, 
3 

B = -yj- 

EX AMPLE. 



L 



P X L 

C X 2fc 



Find width and thickness of a spruce beam 10 feet long, 
when fastened at one end and required to carry, with 8 as factor 
of safety, a load of 1800 pounds at the other end, the thickness 
to be 1 y 2 times the width. 

When the beam is to carry 1800 pounds, with 8 as a factor 
of safety, its breaking load is 8 X 1800 = 14,400 pounds. 



248 STRENGTH OF MATERIALS. 

Solution : 



^_ 14400 X 10 
* 2% X 125 

£=V~bl2 

B = 8 inches in width. 

H—\% X 8" = 12 inches in thickness. 

The weight of the beam itself is not considered in this 
problem. 

To Find the Size of a Beam to Carry a Given Load When 
Also the Weight of the Beam is to be Considered. 

Rule. 

Calculate first the size of beam required to earn* the load, 
then figure what such a beam will weigh and add half of this 
weight to the load, if the beam is fastened at one end and loaded 
at the other, or supported under both ends and loaded at the 
center, but add the whole weight of the beam to the weight of 
the load if the load is distributed along the whole length of the 
beam. Then figure the size of the required beam for this new 
load. 

Example. 

Find width and thickness of a pitch pine beam to carry 
2000 pounds, with 8 as factor of safety, and a span of 27 feet. 
The beam is supported under both ends and loaded at the center ; 
its own weight is also to be taken into consideration. 

Solution : 

Find the constant for pitch pine in Table Xo. 30 to be 150. 
and find the weight of pitch pine in Table Xo. 10 to be 50 pounds 
per cubic foot. When the beam is supported under both ends 
and loaded at the center it is four times as strong as if fastened 
at one end and loaded at the other: therefore, constant 150 is 
multiplied by 4. The load. 2000 pounds, multiplied by 8 as a 
factor of safety, gives 16.000 pounds as breaking load' of the 
beam. 

3 

D ^ 1 16000 X 27 

B = \ 

> 2)4 X 150 X 4 

3 

B — V 320 

B — 6-84" = width, and \% X 6.84" = 10.26" = thickness. 

The area is 6.84 X 10.26 = 70 square inches : the weight per 

foot is 70 times 50 divided by 144. which equals 24.3 pounds, say 25 

pounds. The weight of the' beam is 25 X 27 = 675 pounds. This 



STRENGTH OF MATERIALS. 249 

weight is distributed along the whole beam and, therefore, it does 
not have any more effect than if half of it, or 337^ pounds, 
was placed at the center, but as the beam is to be calculated with 
8 as factor of safety, the weight allowed for the beam must be 
337 x /t X 8 = 2700 pounds. Thus, adding this weight to 16,000 
pounds gives 18,700 pounds ; this new weight is used for calcu- 
ing the size of the required beam. 



_ / 18700 X 27 



2X X 150 X 4 

£ = Vlm 

B = 7.2 inches = width, and W 2 X 7.2" = 10.8 inches, 
thickness. 

This, of course is also a little too small, as only the weight 
of a beam 6.84 inches by 10.25 inches is taken into account, but 
if more exactness should be required the weight of this new 
beam may be calculated and the whole figured over again, 
and the result will be closer. This operation may be repeated 
as many times as is wished, and the result will each time be 
closer and closer, but never exact; but for all practical purposes 
one calculation, as shown in this example, is sufficient. 

Example 2. 

Find width and thickness of a spruce beam to carry 4200 
pounds distributed along its whole length. The span is 24 feet ; 
use 10 as factor of safety, and also allow for the weight of beam. 
The thickness of the beam is to be 1)4 times its width. 

Solution : 

3 

J 4200 X 24 X 10 

> 2X X 125 X 8 
3 

B — V 448 

B = 7.65 inches, and Jf= 11.48 inches. 

w • us. t u 7 - 65 x n - 48 X 24 X 32 . „ , 

Weight of beam = — - = 468 pounds. 

144 

Adding ten times the weight of the beam to ten times the 
weight to be supported, gives 46,680 pounds. 



_ / 466 



2% X 125 X 8 



3 

B — s/ 497.9 

B = 7.93 inches, and H — l l / 2 B= 11.9 inches, or prac- 
tically, a beam 8 inches by 12 inches is required. 



250 STRENGTH OF MATERIALS. 

Crushing and Shearing Load of Beams Crosswise on the 

Fiber. 

Too much crushing load must not be allowed at the ends of 
the beams where they rest on their supports, as all kinds of 
wood has comparatively low crushing strength when the load is 
acting crosswise on the fiber. 

Approximately, the average ultimate crushing strength of 
wood, crosswise of the fiber, is as follows : — 

White oak, 2000 pounds per square inch. 

Pitch pine, 1400 pounds per square inch. 

Chestnut, 900 pounds per square inch. 

Spruce and pine, 500 to 1000 pounds per square inch. 

Hemlock, 500 to 800 pounds per square inch. 

The safe load may he from one-tenth to one-fifth of the 
ultimate crushing load. When the wood is green or water- 
soaked, its crushing strength is less than is given above. 

Example. 

How much bearing surface must be allowed under each 
end of the beam mentioned in Example 2, providing it also 
has 10 as a factor of safety ? The crushing strength of spruce 
crosswise on the fiber is 500 pounds, and using 10 as factor of 
safety, the load allowed per square inch must be only 50 pounds. 
The beam is 8 inches wide, and half of 46S5 pounds is sup- 
ported at each end ; thus the length of bearing required under 

2342 
each end will be _ x t, = 5.85 inches. Thus, the least bearing 

allowable should be about 6 inches long. 

When beams are heavily loaded and resting on posts, or 
have supports of small area, either hardwood slabs or cast-iron 
plates should be placed under their ends, in order to obtain 
sufficient bearing surface for the soft wood. 

The same care must be exercised when a beam is loaded at 
one point; the bearing surface under the load should at least 
be as long as the bearing surface of both ends added together. 

Short beams are liable to break from shearing at the point 
of support, especially when loaded throughout their whole length 
to the limit of their transverse strength. 

The ultimate shearing strength for spruce, crosswise of the 
fiber, is 3000 pounds per square inch (see page 273). Safe load 
may be 300 pounds per square inch. 

In the above example the beam is 8" X 12" = 96 square 
inches, and its center load is 46S5 pounds, or 2342^ pounds at 

23421/ 

each end. The shearing stress is 9 g 2 = 24.6 pounds per 
square inch. Hence, the factor of safety against shearing is 
about 100, and there is not the least danger that this beam will 
give way under shearing; but such is not always the result. 



STRENGTH OF MATERIALS. 25 1 

Round Wooden Beams. 

A round beam has 0.589 times the strength of a square 
beam of same length and material, when the diameter is equal 
to the side of the square beam. The area of a square beam 
compared to the area of the round beam is as 0.7854 to 1 ; there- 
fore it might seem as if that also should be the proportion be- 
tween their strength, which is the case for tensile, crush- 
ing and shearing strength, but not for transverse strength 
or for deflection, because the material is not applied to such 
advantage in the round beam as it is in the square one. All 
preceding formulas for transverse strength of square beams may 
also be used for round beams if only constant C is multiplied 
by 0.589, or, say, 0.6. 

Thus, the formula for a round beam fastened at one end 
and loaded at the other will be : 

0.6 C X Z> 3 
P = —L 

Note. — In a round beam, of course, it will be D z instead of 
ff 2 b for a rectangular one. 

Example. 

Find the load in pounds which will break a spruce beam 12 
feet long and 6 inches in diameter when supported under both 
ends and loaded at the center. (Find constant C in Table No. 30.) 

Solution : 

4 X 0.6 C X D* 



P 



P = 



L 
4 X 0.6 X 125 X 6 X 6 X 6 



12 
P = 5400 pounds. 

To Calculate the Size of Round Beams to Carry a Given 
Load When Span is Known. 

Where the load and span are known, the diameter of the 
beam is calculated, when fastened at one end and loaded at the 
other, by the formula : 



-r. . IP X L X factor of safety 

Rule. 

Multiply together the load in pounds, factor of safety and 
length of span in feet, divide this product by six-tenths of the 



252 



STRENGTH OF MATERIALS. 



constant in Table No. 30, and the cube root of this quotient is 
the diameter of the beam. 

Example. 

A round spruce beam is fastened into a wall, and is to 
carry 1200 pounds on the free end projecting 4 feet from the 
wall, with 8 as a factor of safety, the weight of the beam not to 
be considered. Find diameter of beam. 



Solution : 



D 



3 
3 

=4 



1200 X 4 X 8 
0.6 X 125 



38400 



75 

3 

D — V 512 

D = 8 inches diameter. 



Load Concentrated at Any Point, Not at the Center of 

a Beam. 

If a beam is supported at both ends and loaded anywhere 
between the supports but not at the center (see Fig. 25), it will 
carry more load than if it was loaded at the center. With 
regard to breaking, the carrying capacity is inversely as the 
square of half the beam to the product of the short and the long 
ends between the load and the support. For instance, a beam 
10 feet long is of such size that when it is supported under 
both ends and loaded at the center it will carry 1400 pounds. 
How many pounds will the same beam carry if loaded 3 feet 
from one end and 7 feet from the other ? 



Solution 



X = 



X = 



1400 X 5 2 

7X3 

1400 X 25 

21 



X — 1666% pounds. 
If weight of beam is also in- 
cluded in its center-breaking-load, 
the formula will be : 



FIG. 25. 



P X = (P X 



F 2 



) -y 2 w 




a X & 
P == Breaking load (including 

weight of beam) if applied at the center in pounds. 

p=z Half the length of the span, 



STRENGTH OF MATERIALS. 



253 



IV = Weight of beam. 

P\ = Breaking load applied at n. 

T j t^. >, Load at « X distance £ 

Load on Pier A = 

span 

T j TV „ Load at « X distance « 

Load on Pier B = — 

span 

Beams Loaded at Several Places. 

Fig. 26. 




When a beam is loaded at several places the equivalent 
center load and the load on each support may be calculated as 
shown in the following example: (See Fig. 26). 

The equivalent center load for a — 4 X 20 X 1000 

12 X 12 

The equivalent center load for b = 10 X 14 X 800 

12 X 12 



555.6 lbs. 

777.8 lbs. 



The equivalent center load for c — 16 X 8 X 900 = 800 lbs. 

12 X 12 

The equivalent center load for d = 19 X 5 X 300 = 197.9 lbs. 

12 X 12 

The equivalent center load for loads a, b, c and d is 2331.3 
pounds. 

The load on Pier A — 

(5 X 300) + (8 X 900) + (14 X 800) + (20 X 1000 ) _, im2 y {hs 

24 
The load on Pier B = 

(4 X 1000) + (10 X 800) + (16 X 900) + (19 X 300) _ 1387 T/ lbs 

24 
Note. — The sum of the load on supports A and B is always 
equal to the sum of all the loads ; therefore, by subtracting the. 



254 



STRENGTH OF MATERIALS. 



calculated load on B from the total load the load on A is ob- 
tained. By subtracting the calculated load at A from the total 
load, the load on B is obtained. 

To each load as calculated above for each support also add 
half the weight of the beam. 

To Figure Sizes of Beams When Placed in an Inclined 

Position. 



Fig. 23. 




Figure all calculations concerning 
the transverse strength from the dis- 
tance S, and leave the length L out of 
consideration. If the distance S cannot 
be obtained by measurement it may 
be found by multiplying L by cosine of 
angle a. 



DEFLECTION IN BEAMS WHEN LOADED 
TRANSVERSELY. 

Experiments and theory both prove that if the span is 
increased and the width of the beam increased in the same pro- 
portion the transverse strength of the beam is unchanged ; but 
such is not the case with its stiffness. If a beam is to have the 
same stiffness its depth must be increased in the same ratio as 
the span, providing the width is unchanged. Within the 
elastic limit of the beam the deflection is directly proportional 
to the load; that is, half the load produces half the deflection, 
out doubling the load will double the deflection. 

Deflection is proportional to the cube of the span ; that is, 
with twice the length of span the same load will, when the other 
dimensions of the beam are unchanged, produce eight times as 
much deflection. 

Deflection is inversely as the cube of the depth (thickness) 
of the beam. For instance, if the depth of a beam is doubled 
but the length of span and the width of beam is unchanged, the 
same load will produce only one-eighth as much deflection. 
Deflection is inversely as the width of the beam ; for instance, 
when a beam is twice as wide as another beam of the same 
material but all the other dimensions are unchanged, the same 
load will produce only half as much deflection. 

The deflection in a beam caused by various modes of load- 
ing is calculated by the following formulas : — 

For beams laid in a horizontal position and loaded trans- 
versely, fastened at one end and loaded at the other ; (See Fig. 6). 



$ = 



P X I* 
3XE X / 



STRENGTH OF MATERIALS. 255 

For beams laid in a horizontal position, fastened atone end 
and loaded thoroughout the whole length : (See Fig. 7.) 

S = P X L * 

$ X E X f 

For beams laid in a horizontal position, supported under both 
ends and loaded at the center: (See Fig. 8). 

s _ P X L s 
4SXEXI 

This formula may be transposed and used to calculate 
modulus of elasticity from the results obtained when specimens 
are tested for transverse stiffness. Deflection should be care- 
fully measured but the specimen must not be bent beyond its 
elastic limit; the modulus of elasticity is calculated by the trans- 
posed formula : 

E _ P XI? 
48 X S X I 
For a square specimen / is (side of beam) 4 divided by 12. 
(See moment of inertia, page 237). 

(Also see rule for calculating modulus of elasticity, 
page 265). 

For beams laid in a horizontal position, supported under both 
ends and loaded uniformly throughout their whole length : (See 
Fig. 9). 

S= 5 X P x LS 
384 X E X I 

For beams laid in a horizontal position, fixed at both ends, 
and loaded at the center : (See Fig. 10). 

s- PxL * 



192XEXI 



For beams laid in a horizontal position, fixed at both ends and 
loaded uniformly throughout their whole length : (See Fig. 11). 

6 ._ pxu 



384 X E X I 

In these formulas the definitions of the letters are : 
S = Deflection in inches. 
P = Load in pounds. 
L = Length of span in inches. 

E = Modulus of elasticity in pounds per square inch. 
/= Rectangular moment of inertia. (See pages 237-238). 

These formulas are applicable to any shape of section 
or material, when the load is within the elastic limit. 



256 STRENGTH OF MATERIALS. 

For beams of symmetrical section it is more convenient to 
use the following equally correct but more practical formulas, by 
which the deflection is calculated directly from the size of the 
beam by simply using a constant obtained by experiment and 
reduced by calculation to a unit beam one foot long and one 
inch square, thus avoiding both the use of the modulus of elas- 
ticity and the moment of inertia. 

When beams are supported under both ends and loaded at 
the center, and the weight of the beam itself is not considered, 
the following formulas may be used for solid rectangular beams 
laid in a horizontal position : 

c _ Z 3 X P X c s 

S H*XB L = J^_x_B_XS 

* P* c 



H—^\ L*X B Xc c __ S X H* X B 



=v 



S X B Z 3 X P 



B = ZB X P X c P == H * X B X S 

SX H* Z 3 X c 

S = Deflection in inches. 

Jf= Thickness of beam in inches. 

B = Width of beam in inches. 

Z = Length of beam in feet. 

P = Load in pounds. 

c = Constant obtained by experiment, and is the deflec- 
tion, in fractions of an inch, which a beam one foot long and 
one inch square will have if supported under both ends and 
loaded at the center; the average value for this constant is 
given in Table No. 31. 

For any other mode of loading, see rules and explanations 
on page 261. 

In previous formulas and rules, the weight of the beam 
itself was not considered. The deflection in a beam caused by 
its own weight when it is of rectangular shape and uniform 
size, and laid in a horizontal position, is obtained by the 
formula, 

Z 3 X H W X c 



S 



H*X B 



When both the weight and the load are to be considered, 
the deflection in a solid rectangular beam laid in a horizontal 
position, supported under both ends and loaded at the center, is 
calculated by the formula, 

i*{p+nmc 

H*X B 



STRENGTH OF MATERIALS. 



257 



6* = Deflection in inches. 

L = Length of span in feet. 

P = Load in pounds. 

W=z Weight of beam in pounds. 

c = Constant obtained by experiments, and is the deflec- 
tion in fractions of an inch, which a beam one foot long and 
one inch square will have if supported under both ends and 
loaded at the center, and may be found in Table No. 31. 

J7= Thickness of beam in inches. 

B = Width of beam in inches. 

Rule. 

To the load add five-eighths of the weight of beam, mul- 
tiply this by the cube of the length of the span in feet, and 
multiply by a constant from Table No. 31. Divide this product 
by the product of the cube of the thickness and the width of 
the beam ; the quotient is the deflection in inches. 

The deflection in a beam supported under both ends and 
loaded evenly throughout is five-eighths of that of a beam 
supported under both ends and loaded at the center. Therefore, 
in_ the following formulas, the weight of the beam itself is multi- 
plied by five-eighths to reduce the effect of the weight of the 
beam to the equivalent of a load placed at its center. 

FOR SOLID SQUARE BEAMS. 



.9 ^ c(P±X W)L* 




Fig. 28 



FOR SOLID RECTANGULAR BEAMS. 



s _ c{P+ % W)L * 
B H* 




V B "i 



FOR HOLLOW SQUARE BEAMS. 



H± — h* 



Ital 



-7H 



FIG. 30 



25* 



STRENGTH OF MATERIALS. 



FOR HOLLOW RECTANGULAR BEAMS. 

JtM 



7? HZ h h* i. %. 



B H* — b k 



I I 






Li 



V/////////M 



-B- 



"** FIG. 31 



FOR I BEAMS. 



r 



X — c(P + H W) Z 3 i 

B H* — 2bh* 



*Ub\ 



^ | I Fig. 32 



—- B— 

FOR SOLID ROUND BEAMS. 



<?_i.i c(p+ y 8 W) Z 3 

■* ~m 




FIG. 33 



FOR HOLLOW ROUND BEAMS. 



S -. Wc{P + ft IV) Z 3 
Z> 4 — d* 




Fig. 34 



FOR SOLID ELLIPTICAL OR OVAL BEAMS. 

Z>i x z> 3 Wll F,G ' 



35 



-Br 

FOR HOLLOW ELLIPTICAL OR OVAL BEAMS. 



o_ 1.7g(i p +# ^)Z 3 ] 
Z>i Z> 3 —dxd* 1 




STRENGTH OF MATERIALS. 



259 



6" = Deflection in inches. 

L — Length of span in feet. 

P = Load in pounds. 

W=- Weight of beam in pounds. 

c = Constant obtained from experiments, or may be ob- 
tained from Table No. 31. 

For meaning of the other letters, see figure opposite each 
formula. 

A round beam equal in diameter to the side of a square 
beam will deflect 1.698 times as much, and for convenience, 
when the deflection of a square or a rectangular beam, whether 
solid or hollow, is known, it may be multiplied by 1.7, and the 
product is the deflection of a corresponding round, oval, or 
elliptical beam of the same material and diameter and laid in 
the same relative position and loaded in the same manner as 
the calculated beam. It is well to remember that a round or 
elliptical beam weighs a little less than a square or rectangular 
one, when the sides and diameters are equal, and the deflection 
due to its own weight is, therefore, a little less. 



TABLE No. 3 1.— Constant c, 

Giving deflection in inches per pound of load when the beam 
is supported under both ends and loaded at the center. 



Material. 


Constant c. 


Material. 


Constant c. 


Cast steel, 
Wrought iron, 
Machinery steel 
Cast-iron, . 


0.0000143 
0.0000156 
0.C000156 
0.0000288 


Pitch pine, 

Spruce, 

Pine, 


0.00024 
0.00035 
0.00033 



Example. 

A beam 6" X 9" of pitch pine, 10 feet long, supported 
under both ends, is to be loaded at the center with one-tenth 
of its breaking load. Find the load and deflection. 

Solution : 

9 2 X 6 X 4 X 150 



P = 



P — 



10 X 10 

291600 



= 2916 pounds. 



100 
Deflection will be, 

S ~ 1Q3 X 2916 X 0.00024 _ 699.84 _ 16 inch 



9 3 X 6 



4374 



260 STRENGTH OF MATERIALS. 

Therefore, if this beam had been curved 0.16 inch upward, 
by increasing its thickness on the upper side, it would have been 
straight after the load was applied* 

In this example the weight of the beam itself is not 
considered either in figuring the strength or the deflection, 
because the beam is comparatively short in proportion to its 
width and thickness. The weight of the beam itself will only 
be about 200 pounds, and this will be of no account in propor- 
tion to the load that the beam will carry, with 10 as a factor of 
safety. The weight of the beam will increase its deflection cnly 
0.006 inch. In such a beam the danger is probably greater from 
crushing of the ends at the supports, if it has not enough bear- 
ing surface. In long beams the weight of the beam must not 
be neglected, either in calculating safe load or in calculating 
deflection. 

Example 2. 

A round bar of wrought iron is 5 feet long and 3 inches in 
diameter, and loaded at the center with 800 pounds. How much 
will it deflect ? A round bar of iron 3 inches in diameter and 
5 feet long weighs 119 pounds. (See table of weights of iron, 
page 143.) 

Solution : 

c _ 5 3 X (800 4- % X 119) X 1.7 X 0.0000156 
^ _ 3* 

S = 0.0359 inch. 

Thus, such a shaft loaded with 800 pounds will deflect T ff ¥ 
of an inch in the length of 5 feet, or 60 inches. If the deflec- 
tion must not exceed yzqq of the span (see page 266), then the 
greatest allowable deflection for this span would be 0.04 inch, 
and the calculated deflection is within this limit. 

Note. — ^Vo of the span is equal to a deflection of 0.008 
inch per foot of length. 

Example 2. 

A shaft of machinery steel, 11 inches in diameter and 6 feet 
between bearings, carries in the center a 12-ton fly wheel. How 
much deflection will the weight of the fly wheel cause? 

Note. — Such shafts are usually considered as a beam sup- 
ported under both ends. (See formula for deflection in solid 
round beams, page 258.) 

Solution : 

12 tons = 24,000 pounds. (Weight of shaft is not taken 
into consideration.) 

* This is a thing, frequently done in practice. 



s = 



STRENGTH OF MATERIALS. 26 1 

l a p x \nc 

LP 
6 3 X 24000 X 1.7 X 0.0000156 



ll 4 
216 X 24000 X 0.00002652 



14641 
S = 0.00939 inches. 
Thus, the calculated deflection caused by the fly wheel is a 
little less than T £„ of an inch. The deflection per foot of span 
will be 0--0 P_9_ 3 9 which equals 0.001565 inch. 

Example 3. 

Calculate the deflection of shaft mentioned in the previous 
example, when both the weight of fly wheel and the weight of 
shaft are to be considered. 

Solution : 

6 3 X (24000 + H X 1920) X 1.7 X 0.0000156 



S = 

s = 



ll 4 

216 X 25200 X 0.00002652 



14641 
6" ='0.00986 inch. 

Practically, the deflection is likely to be a little less than 
what is figured in the two previous examples, because if the 
hub of the fly wheel fits well on the shaft, it will stiffen it some. 
(It is a good practice to make such shafts a little larger in 
diameter in the place where the hub of the wheel is keyed on ; 
this enlargement will then compensate for what the shaft is 
weakened by cutting the key-way.) 

The weight of the shaft may be obtained by considering a 
cubic foot of machinery steel to weigh 485 pounds, and a shaft 
11 inches in diameter will then weigh 320.1 pounds per foot in 
length, and 6 feet will weigh 1920 pounds. Multiplying this by 
Y% gives 1200 pounds, to be added to the weight of the fly wheel, 
which gives 25,200 pounds. The weight of the shaft may also 
be found in the table of weight of round iron, page 144. 

To Calculate Deflection in Beams Under Different Modes 
of Support and Load. 

Constant c in Table No. 31 is the deflection in fractions of 
an inch per pound of load when a beam one foot long and one 
inch square is supported under both ends and loaded at the 
center, and when this constant for any given material is known, 
the deflection for beams subjected to other modes of fastening 
and loads may be calculated thus : 

For beams supported under both ends with the load dis- 
tributed evenly throughout their whole length, multiply c by ji . 



262 STRENGTH OF MATERIALS. 

For beams fixed at both ends and loaded at the center, 
multiply c by %. 

For beams fixed at both ends with the load distributed 
evenly throughout their whole length, multiply c by yi. 

For beams fixed at one end and loaded at the other, mul- 
tiply c by 16. 

For beams fixed at one end with load distributed evenly 
throughout their whole length, multiply c by 6. 

Example. 

A square, hollow beam of cast-iron, 8 inches outside and 6 
inches inside diameter, and 9-foot span, supported under both 
ends, is loaded at the center with 8000 pounds. How much 
will it deflect ? 

Solution : 

Weight of beam = 9 X 12 X (8 2 — 6 2 ) X 0.26 = 786 pounds. 



s _ 9 3 X (8000 + s/g X 786) X 0.0000288 

8 4 — 6 4 
S _ T29 X 8492 X 0.0000288 

4096 — 1296 
s _ 178.291 

2800 
S = 0.064 inches. 



Example. 

How much would this same beam deflect if the load had 
been distributed evenly throughout its whole span ? 

Solution : 

s _ L* (P + W) y 8 c 
Zfi — d* 

~ _ 9 3 X 8786 X H X 0.0000288 
"* S^-6* 

_y_ 115.289 

~ 2800 
S = 0.041 inch. 

Example. 

A round cast-iron beam of 7 inches outside and 5 inches 
inside diameter is 4 feet between supports, with a load of 2000 
pounds distributed evenly throughout its span. How much will 
it deflect, the weight of beam itself not being considered in the 
calculation ? 



STRENGTH OF MATERIALS. 



263 



Solution 
6- 



4 3 X 2000 X 0.0000288 X 1.7 X % 



74 _ 54 

t y_ 256 X 2000 X 0.0000288 X 1.7 X ^ 

1776 
S = 0.0085 inch. 

In this example, 1.7 is used as a multiplier because the 
beam is round, and y% because the load is distributed evenly 
throughout the length of the span. 

Example. 

A fly wheel weighing 800 pounds is carried on the free end 
of a 3-inch shaft, 1 foot from the bearing. How much will the 
shaft deflect? 

This is the same as a round beam loaded at one end and 
fastened at the other; therefore, constant c is multiplied by 
16 X 1.7. 

Solution : 

Z 3 P\.n c X 16 



s 



s = 



1 X 800 X 1.7 X 0.0000156 X 16 



3* 



6* = 0.0042 inch. 



FIG. 37 



Previous calculations for breaking load and also for 
deflection are based upon a dead load slowly applied and not 
exposed to jar and vibrations. If the load is applied suddenly 
it will have greater effect toward breaking the beam than if 
applied slowly. For instance, imagine a load having its whole 
weight hanging on a rope, like Fig. 37, just touching the beam 
but not actually resting upon it. 
If that rope was cut off suddenly 
this load would produce twice as 
much effect toward breaking the 
beam and would cause twice as 
much deflection as if it was loaded 
on gradually. A railroad train 
running over a bridge will, for the 
same reason, strain the bridge 
more when running fast than it 
would if running slow. 




To Find a Suitable 



Size of Beam 
Deflection. 



for a Given Limit oi 



For a square beam supported under both ends and loaded 
at the center, use the formula : 



264 STRENGTH OF MATERIALS. 



VTZ p - 


A round beam supported under both ends and loaded at the 
center may be calculated by the formula : 

4 

T 3 p 1 _ - 

Diameter of beam = V — 

A rectangular beam supported under both ends and loaded 

at the center, and having its depth 1)4 times its width, may be 

calculated by the formula : 

4 

3 U" P c 
Depth or thickness of beam = a " 

L = Length of span in feet. 
P = Center load in pounds. 
5 = Given deflection in inches. 
c = Constant given in Table Xo. 31. 

Note. — These three formulas are only approximate, as 
the weight of the beam itself is not considered : but if 
necessary, after the size of beam is obtained- its weight may 
be calculated and five-eighths of i: added to the center load, 
P: and using the same formula again, another beam may 
be calculated for this new center-load, and this new calculatioD 
will give a beam only a mere trifle too small. Constants in 
Table No. 31 are for beams supported under both ends and 
loaded at the center. For any other mode of loading or fasten- 
ing, constant c must be multiplied according to rules on 

To Find the Constant for Deflection. 

If experiments are made upon rectangular beams, use 
formula. 

■g H» B 

l s (p+ %w) 

Example. 

Calculate the constant c, or deflection in inches per pound 
of load, for a beam of 1 foot span and 1 inch square, supported 
under both ends and loaded at the center, vhen experiments are 
made upon a pitch pine beam 40 feet long. 12'' by S". weighing 
1200 pounds and deflecting 1% inches for a center-load of 500 
pounds. 

Solution : 

_ 1.5 X lg X B 

C ~ 40 s X (500 -f h X 1200) 
c — 0.000259 inch. 



STRENGTH OF MATERIALS. 265 

Modulus of Elasticity Calculated from the Transverse 
Deflection in a Beam. 

When experiments are made upon rectangular beams sup- 
ported under both ends and loaded at the center, the modulus 
of elasticity may be calculated by the formula, 

E - L* {P + H W ) 
4S T*B 

E = Modulus of elasticity. 
L = Length of span in inches (not in feet). 
P = Load in pounds. 
W = Weight of beam in pounds. 
6*= Deflection of beam in inches. 
T= Thickness of beam in inches. 
B = Width of beam in inches. 

Example. 

Calculate the modulus of elasticity for a pitch pine rec- 
tangular beam weighing 1200 pounds, 40 feet span, and 12" by 
8", deflecting 1)4 inches for a center-load of 500 pounds. 
(This beam and conditions are the same as mentioned in 
the previous example for calculating constants.) 



Solution : 
E = 

E = 



480 3 X (500+^X1200) 
4 x iy 2 X 12 3 X 8 

138240000000 



82944 
E = 1,666,666 pounds per square inch. 

This deflection was obtained by actual experiments on a 
pitch pine beam of the dimensions given, and the calculated 
modulus of elasticity agrees fairly well with what is usually 
given by different authorities in tables of modulus of elasticity. 
When experimenting it is necessary to take the average of 
several experiments with different loads and to try the beam by 
turning it upside down, as very frequently it will then deflect a 
different amount under the same load. Care should be taken 
that the load is not so great as to strain the beam beyond 
its elastic limit. As long as the deflection increases regularly in 
proportion to the load, it is a sign that the elastic limit is not 
reached. It is very difficult to ascertain exactly when deflection 
will commence to increase faster than the load, because material 
is never so homogeneous but that the deflection will be more or 
less irregular, although by care and patience fairly good re- 
suits may be obtained. 



266 STRENGTH OF MATERIALS. 



Allowable Deflection. 

The greatest amount of deflection which may be allowed in 
different kinds of construction can only be determined by prac- 
tical experience and good judgment of the designer. As a rule, 
in iron work the deflection is seldom allowed to exceed y^Vo °f the 
span, which is equal to T ^. or 0.008 inch per foot of span. Line 
shaftings are sometimes allowed to deflect T ^ „ of the distance 
between hangers which is equal to 0.01 inch per foot of span, 
but head shafts carrying large pulleys are generally not allowed 
to deflect more than 0.005 per foot of span. 

In woodwork, considerable more deflection is allowed than 
in iron structures. Beams in houses are frequently allowed to 
deflect 3^-q, or even t \q of the span; this is equal to 0.024 to 
0.025 inch*, per foot of span. Woodwork to which machinery is 
to be fastened must never be allowed to deflect so much. Such 
woodwork must always be so stiff that it supports the machinery, 
and not vice versa; for instance, in beams or posts by which 
hangers and shafting are supported, it is not all-sufficient that 
they are strong enough, but they must also always be stiff enough. 

In factories it is very important that floor beams as well 
as beams supporting heavy shafting have sufficient stiff- 
ness as well as strength. Floors in factories are frequendy 
loaded up to 300 pounds per square foot of surface. For floors 
in public buildings, which are never loaded with more than the 
weight of the people who can get room, the load will hardly ex- 
ceed 150 pounds per square foot of surface. Floors in tene- 
ment houses are seldom loaded more than 60 pounds per square 
foot. 

Slate roofs weigh about 8.5 pounds per square foot of 
surface. Snow may be reckoned, when newly fallen, to weigh 
5 to 15 pounds per cubic foot, and when saturated with water 
it may weigh 40 to 50 pounds per cubic foot. Usual practice is 
to allow 15 to 20 pounds per square foot for snow and wind on 
roofs. 

TORSIONAL STRENGTH. 

The fundamental formula for torsional strength is, 

Pm = S — 

a 

Pm = Twisting moment, and is the product of the length 
of the arm, m, in inches and the force. P, in pounds. 

S = Constant computed from experiments, and is some- 
times called the modulus of torsion; its value usually agrees 
closely to the ultimate shearing strength per square inch of the 
material. 

/ = Polar moment of inertia (see page 297). 

a = The distance in inches from the axis about which the 
twisting occurs to the most remote part of the cross section. 

* 0.025 is one-fortieth inch per foot of span. 



STRENGTH OF MATERIALS. 



267 



Example 1. 

A round cast-iron bar 3 inches in diameter, is exposed to 
torsional stress; the length of the lever, w, is 18 inches. Find 
the breaking force, P, in pounds when the modulus of torsion 
for cast-iron is taken as 25,000 pounds. 

Polar moment of inertia for a circle of diameter, d, is, 

d*TT 

32 



The distance a = )4 d. 
9snnn \ 



25000 X 3 4 X 3.1416 X T \ 



P — 



is x \y 2 

25000 X 81 X 3.1416 
27 X 32 



P = 7363^ pounds. 

The advantage of the above formula is that it may be 
used for any form of section, because it takes in the polar 
moment of inertia of the section; but it is seldom that 
calculations of torsional strength are required for other than 
beams of round or square section, either hollow or solid, and 
the strength of such beams may be more conveniently cal- 
culated in an equally correct, but easier way, obviating the use 
of both the polar moment of inertia and the modulus of torsion, 
by reasoning thus : 

Consider two shafts, a and b, Fig 38. Shaft a has twice 
the diameter of shaft b, and consequently four times the area ; 
therefore it has, so to say, four times as many fibers to resist 
the stress, and for this reason it must be four times as strong 
as shaft b ; but, further, the outside fibers in a are twice as far 
from the center, therefore the fibers in shaft a must also have 
on an average twice the advantage over the fibers in shaft b to 
resist the twisting effort of any load exerting a twisting stress, 
and for this reason shaft a must have twice the strength of b ; 
and taking these two reasons together, shaft a must, conse- 
quently, be eight times as strong as shaft b, in resisting tor- 
sional stress. 

Thus, the strength of a solid shaft 
increases as the cube of the diameter. 
Shaft a is twice as large in diameter 
as shaft b, and is, therefore, eight times 
as strong as b, because 2 3 = 8. If 
shaft a had been three times as large 
as b it would have been 27 times as 
strong, because 3 3 = 27 ; if shaft a had 
been four times as large as h, it would have been 64 times as 
strong, because 4 3 = 64, 




268 



STRENGTH OF MATERIALS. 



Therefore, if the constant corresponding to a load, which, 
applied to an arm one foot long will twist off or destroy a bar 
one inch in diameter, is found, the breaking load for any round 
shaft of the same material when under torsional stress may be 
easily calculated. The torsional strength (but not the torsional 
deflection in degrees) is independent of the length of the shaft. 
The strength depends only upon the kind and the amount of 
material, and the form of cross-section. A square shaft having 
its sides equal to the diameter of a round shaft will have ap- 
proximately 20% more strength than the round one, but it will 
take nearly 28% more material. A square shaft of the same 
area as a round shaft has approximately 15% less torsional 
strength than the round one. 

Thus : 

Formulas for torsional strength relating to solid round 
shafts will be : 



P = 



D*c 



m 



m 



P 



D 



=4 



P m 



Pm 



P = Breaking load in pounds. 

D = Diameter of shaft in inches. 

m = Length in feet of the arm on which load P is acting. 

c = Constant, and it is the load in pounds which, when 
applied to an arm one foot long, will twist off or destroy a round 
bar one inch in diameter. This constant is obtained from ex- 
periments, and is given in Table No. 32. 

Rule. — Multiply the cube of the diameter in inches by the 
constant c, in pounds, divide this product by the length of the 
lever m,'m feet, and the quotient is the breaking load in pounds. 



TABLE No. 32.— Constant c. 

The ultimate torsional strength in pounds of a round beam 
one inch in diameter, when load is acting at the end of a lever 
one foot long. 



Material. 


Very Good. Medium Good. 


Poor. 


Cast Steel 

Machinery Steel* . . . 
Wrought Iron .... 
Cast-iron 


2,000 

1,200 

800 

525 


1,000 

1,100 

580 

450 


600 
700 
500 
350 



* Machinery steel or wrought iron may not actually break at this load, but it 
will deflect and yield so it will become useless. 



strength of materials. 269 

Example 1. 

A wrought iron shaft is eight inches in diameter, and the 
force acts upon a lever two feet long. How much force must 
be applied in order to twist off or to destroy the shaft? 

Solution : 

p = 8* X 580 = 512 X 580 = 1484g0 pounds 

Example 2. 

A force of 870 pounds is acting with a leverage of four feet 
in twisting a wrought iron shaft. What must be the diameter 
of the shaft in order to resist the twisting stress, with 10 as a 
factor of safety? 

Solution : 

3 

D _ J P m X 10 

* c 

3 



= #' 



^ /870 X 4 X 10 



580 

3 

D = V 60 = 3.914, or, practically, a 4-inch shaft. 

Note. — Ten is used as a multiplier of the twisting moment, 
P m, because 10 is the factor of safety. Constant 580 is taken 
from Table No. 32. 

Example 3. 

A round bar of cast-iron four inches in diameter is to be 
twisted off by a force of 3200 pounds. How long a leverage is 
necessary ? (c for cast-iron, in Table No. 32, is 450). 

Solution : 

Z> 3 c 



m = 



P 



m = ^ X 45 ° - 64 X 45 ° =9 feet long. 
3200 3200 & 

Example 4. 

Experiments are made upon a cast-iron round bar 2 inches 
in diameter with a leverage of h% feet ; the bar is twisted off at 
a force of 832 pounds. Calculate constant c, or the force in 
pounds if acting with a leverage of one foot, which will break a 
round bar of the same material one inch in diameter. 

Solution : 

P m 

, = 882X5X = 4868 = 546 pound ^ 



270 STRENGTH OF MATERIALS. 

Hollow Round Shafts. 

In proportion to the amount of material used, around hollow 
shaft has more torsional strength than a solid shaft of the same 
diameter. This is because the fibers in any shaft exposed to 
twisting stress only offer resistance to the load in proportion to 
their stretch. Therefore, the fibers near the center are always 
in position to offer less resistance than the fibers more remote 
from the center. 

The formula for torsional strength in round hollow shafts 
will be : 



V D X m J 



P = Ultimate breaking load in pounds applied at a leverage 
of m feet. 

D = Outside diameter of shaft in inches. 

d= Inside diameter of shaft in inches. 
vt = Length of lever in feet. 

c = Constant (same as for a solid shaft). 



Square Beams Exposed to Torsional Stress. 

The theoretical formula for twisting strength (on page 266) 
will apply to square as well as round beams. The proportional 
strength between a round and a square beam may, therefore, be 
compared by using that formula. Let S represent the side of a 
square beam and the polar moment of inertia is \ S*. 

The distance from the center of the beam to the most 
remote fiber in a square beam is S V }4, and, dividing the polar 
moment of inertia by this distance, we have, 



1 94 

6 ° 



= 0.23 S 3 



Let D represent the diameter of a round beam. The 

polar moment of inertia is = 0.098 D* 

F 32 

The distance from the center to the most remote fiber in the 

round beam is y 2 D. Dividing the polar moment of inertia by 

this distance, we have °- 098 ^ — 0.196 D z 

Vz D 

Suppose, now, that S and D are equal, for instance, 
one inch ; the proportion in torsional strength between the two 
beams must be 0.23 divided by 0.196, which equals 1.18. 
Thus, for square beams, use the formulas given for round 
beams, but multiply constant c, in Table No. 32, by 1.2, and 



STRENGTH OF MATERIALS. 27 I 

take the side instead of the diameter. The formula for tor- 
sional strength in a square beam will be : 

p__ (Side) 3 X 1.2 X c 

Length of leverage. 

c == Constant (same as for a round beam). 

P '==■ Load in pounds. 

Side is measured in inches. 

Length of leverage is measured in feet. 

Torsional Deflection. 

The torsional deflection in degrees will increase directly 
with the length of the shaft and the twisting load, and inversely 
as the fourth power of the diameter of the shaft ; therefore, the 
formula for torsional deflection is : 

s _c X m X L X P 

S = Deflection in degrees for the length of the shaft. 

m = Length of lever in feet. 

L = Length of shaft in feet. 

P — Load in pounds. 

D = Diameter of shaft in inches. 

c = Constant obtained from experiments for different 
kinds of material, and is the deflection in degrees for a shaft 
one inch in diameter and one foot long, when loaded with one 
pound on the end of a lever one foot long. 

The author of this book has made experiments on tor- 
sional deflection in wrought iron shafts two inches in diameter. 
The average deflection was \ x / 2 degrees in 10 feet of length, 
when a load of 50 pounds was applied on a lever h% feet long. 
Constant c, as calculated from these experiments, will be 0.00914. 
Using this constant, the formula for torsional deflection for 
wrought iron will be : 

S _ L X m X P X 0.00914 

Machinery steel and wrought iron will deflect about the 
same. Cast-iron will deflect twice as much as wrought iron. A 
square bar will deflect 0.589 times as much as a round bar when 
side and diameter are alike. 

Formula for Torsional Deflection in Hollow Round Shafts. 

~_ LXm X PXc 

D = Outside diameter in inches. 
d— Inside diameter in inches. 

All the other letters have the same meaning as explained 
under formulas for solid shafts. 



272 



STRENGTH OF MATERIALS. 



TABLE No. 33.— Constant c. 

The torsional deflection in degrees per foot of length for a 
shaft of one inch side or diameter when loaded with one pound 
at the end of a lever one foot long : 



Material. 



Machinery Steel 
Wrought Iron . 
Cast-iron .... 

Oak 

Ash 

Pine and Spruce 



Round 


Square 


Section. 


Section. 


0.00914° 


0.00538° 


0.00914° 


0.00538° 


0.018° 


0.0106° 


0.795° 


0.468° 


0.784° 


0.460° 


1.35° 


0.79° 



Example. 

A round bar of wrought iron 16 feet long and 3 inches in 
diameter is fastened at one end and the other is exposed to a 
twisting load of 1000 pounds, acting with 5 feet leverage. How 
many degrees will this load deflect the bar ? 



Solution 



S= 



S= 



16 X 5 X 1000 X 0.00914 



3 4 



731.2 
81 



S= 9 degrees. 

Note. — From Table No. 33, it is seen that only steel and 
wrought iron are suitable for shafts exposed to torsional stress. 
Wrought iron is about twice as good as cast-iron, over 80 times 
better than oak, and about 150 times as good as pine. 



SHEARING STRENGTH. 

Sometimes force may act in such a manner that the 
material is sheared off. For instance, the rivets in a steam 
boiler are exposed to shearing stress (see Fig. 39) when the 
boiler is under steam pressure. 

FIG. 39. 



When holes are punched or bars of 1 'T? ~i 

iron are cut off under punching presses, I vL Jr— * 

the action of the punch in cutting off the ~* •» »»• »- 

material is shearing, and the resistance which the material offers 
is its ultimate shearing strength. The average ultimate shear- 
ing strength of wrought iron is 40,000 pounds per square inch. 



STRENGTH OF MATERIALS. 273 

In cast-iron the ultimate shearing strength is usually between 
20,000 and 80.000 pounds per square inch. In steel the ultimate 
shearing strength will vary from 40,000 to 80,000 pounds per 
square inch. 

The resistance offered to shearing is in proportion to 
the sheared area. Thus, it will take twice as much force 
to punch a hole two inches in diameter through a three-eighths 
inch plate as it would to punch a hole only one inch in diameter 
through the same plate, and it will take four times as much 
force to shear off a one-inch bolt as it would to shear off a one- 
half inch bolt, because the area of a one-inch bolt.is four times 
as large as the area of a one-half inch bolt. 

Example 1. 

How much force is required to shear off a wrought iron 
rivet of one-inch diameter if the shearing strength of the 
wrought iron is 40.000 pounds. 

Solution : 

One-inch diameter = 0.7854 square inches; therefore the 
force required will be 0.7854 X 40,000 = 31,416 pounds. 

Example 2. 

A wrought iron plate is one-quarter of an inch thick and 
the ultimate shearing strength of the iron is 40,000 pounds per 
square inch. How much pressure is required to punch a hole 
three-quarters of an inch in diameter ? 

Solution : 

The circumference of a %-inch circle is 2.356 inches. The 
plate is X _m ch thick; therefore the area of shearing surface, 
2.3562 X % = 0.58905 ; thus, the force required will be 40,000 
X 0.58905 = 23,562 pounds. 



TABLE No. 34.— Shearing Strength Per Square Inch. 



Material. 



Steel 

Wrought Iron Rivets , 

Cast-iron • 

Oak, crosswise . . . 
Oak, lengthwise . . , 
Pitch Pine, crosswise , 
Pitch Pine, lengthwise 
Spruce, crosswise . . 
Spruce, lengthwise . 



Pounds Per Square Inch. 



45,000 to 


75,000 


35,000 to 


55,000 


20,000 to 


30,000 


4,500 to 


5,500 


400 to 


700 


4,000 to 


5,000 


400 to 


600 


3,000 to 


4,000 


300 to 


500 



274 



STRENGTH OF MATERIALS. 



FACTOR OF SAFETY. 

The factor of safety can only be fixed upon by the experi- 
ence and good judgment of the designer. It may vary from 4 to 
40. In a temporary structure, when the greatest possible load to 
which it will be exposed is known, a factor of safety of four 
may be safe enough, but frequendy a greater factor is necessary. 
Different factors of safety are also necessary for different 
materials; a different factor of safety may also be necessary 
in different parts of the same machine. The following Table, 
No. 35. is only offered as a guide in selecting factor of safety- : 

TABLE No. 35.— Factor of Safety. 



Material. 


Dead Load, 
such as build- 
ings contain- 
ing little or no 

machinery. 


Variable Load, 
such as bridges 
and slow- 
running 
machinery. 


Machinery 

in 

General. 


Machinery 

Exposed to 

hard usage, as 

Rolling Mills, 

etc. 


SteeL 

Wrought Iron. 
Cast-iron. 
Brickwork. 
Wood. 


5 

4 
6 

15 


T 

6 

10 

25 
10 


10 
10 
15 
30 

15 


15 
15 
25 

40 

20 



If a structure is exposed to stress alternately in one direc- 
tion and then in another, it is necessary to use a higher factor of 
safety than if it is only exposed to a steady stress one way. A 
comparatively small load, when applied a sufficient number of 
times, may break a structure or a machine, although it does not 
break it the first time. For instance, commence to hammer on 
a bar of cast-iron and it will break after several blows, 
although the last blow need not be any more powerful 
than the first one. It is the same way with anything else: it 
may break in time, although it is strong enough to resist the 
stress at the beginning: therefore, within practical limits, the 
larger the factor of safety the longer time the structure may 
last 






NOTES ON STRENGTH OF HATERIAL. 

In steel, the crushing strength usually exceeds the tensile 
strength, but wrought iron has usually a little more tensile than 
crushing strength, and its shearing strength is about 80 per cent. 
of its tensile strength. Both steel and wrought iron are suitable 
to resist any kind of stress, and compared to other materials 
they are especially adapted for anything exposed to twisting and 
shearing stress. 



STRENGTH OF MATERIALS. 275 

Cast-iron is variable ; it has usually five to six and a-half 
times as much crushing as tensile strength, and when loaded 
transversely it will deflect under the same load nearly twice as 
much as wrought iron. It is especially useful for short pillars 
or anything exposed to crushing stress, where there is little 
danger of breakage by flexure; it is very much less reliable 
when exposed to tensile or torsional stress. 

Wood is not adapted to resist torsion, but is useful to resist 
tensile, crushing and transverse stress, also to resist flexure. 
It has nearly twice as much tensile as crushing strength ,; there- 
fore, it would seem specially well adapted, in all kinds of con- 
struction, to be the member exposed to tensile stress, but where 
wood and iron enter into construction together, iron is 
always used as the member to take the tensile stress and wood 
as the compressive member, because wood has such, low shear- 
ing strength lengthwise with its fibers that, with any kind of 
fastening at the ends, it will tear and split at the holes under 
comparatively little stress; but this difficulty is easily overcome 
when wood is used as the compressive member. Wood has 
comparatively low tensile and crushing strength crosswise on the 
fiber. This is well to remember with beams loaded transversely 
and laid on posts. The beams may be sufficiently strong, but 
under heavy load, if suitable precautions are not taken (see 
page 250) the top of the post may press into the beam, especially 
if the lumber is green. 

Stone has high crushing strength but low tensile strength, 
and, in consequence, very low transverse strength. It is very 
well adapted for foundations when supported and laid in such a 
way that its crushing strength comes into play, but when laid as 
a beam to resist transverse stress it is very unreliable, as it will 
break for a comparatively small load and it may break from a 
blow or jar. 

Brickwork is only suitable for crushing stress, and there is 
great difference in the strength of different kinds of brick. 

In calculating strength and stiffness in any kind of design- 
ing, it should be remembered that it is only possible to deter- 
mine the strength of any material by actual test, and that the 
tabular and constant numbers here given are only an average 
approximate. 



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MECHANICS. 277 

through is 64.4 feet, because the average velocity per second 
must be half of the final velocity; therefore, the average 
velocity is 32.2 feet per second, and, as the time is two seconds 
the space will be G4.4 feet. At the end of three seconds the 
final velocity has increased to 3 X 32.2 = 96.6 feet per second 
and the space fallen through is -2 | S. XS= 144.9 feet, etc. This 
is supposing the body was falling freely in vacuum, but while 
the air will offer a resistance and somewhat reduce the actual, 
motion, the principle is the same. Acceleration due to gravity 
varies but little at different latitudes of the earth. At the 
equator it is calculated to be 32.088 and at the pole 32.253 feet. 
Acceleration due to gravity decreases at higher altitudes,* but 
all these variations on the earth's surface are so small that they 
hardly need to be considered in any calculation concerning 
practical problems in mechanics. 

Velocity. 

The velocity of falling bodies increases at a uniform rate of 
32.2 feet per second ; therefore, when commencing from rest, the 
final velocity in feet per second must be, 

Rule. 

Multiply the time in seconds by 32.2 and the product is the 
final velocity in feet per second ; or, multiply the height of the 
fall in feet by 64.4 and the square root of the product is the 
velocity in feet per second. 

Example. 

What final velocity will a body acquire in a free fall during 
seven seconds ? 

Solution : 

v = 1 X 32.2 = 225.4 feet per second. 

Height of Fall. 

The average velocity per second is always half of the final 
velocity per second. Therefore the space fallen through in a 
given time is found by multiplying half of the final velocity by 
the number of seconds which produced that velocity. Thus, the 
formulas : 

k = — = t0.6v= v 0.5 / = ^- — ^ = t 2 0.b g 
2 2^2 

* Above the surface of the earth the weight of a body is inversely propor- 
tional to the square of its distance from the center of the earth. 

Below the surface of the earth the weight oi a body is directly proportional to 
its distance from the center ot the earth. 



278 MECHANICS. 

Example. 

A fly-wheel has a rim speed of 4S feet per second. From 
what height must a body drop to acquire the same velocity? 

Solution : 

* = if. = J*_ = 23M = 33.78 feet. 
2 g 64.4 644 

Time. 

Rule. 

Divide the space by 16.1, and the square root of the 
quotient is the time : or, divide given velocity by 32.2, and the 
quotient is the time. 



S > U.5 



8 

Example. 

How long a time does it take before a body in a free fall 
acquires a velocity- of 100 feet per second ? 
Solution : 

/ = = — = 3.1 seconds. 

g 32.2 

Distance a Body Drops During the Last Second. 

The space through which a body will drop in the last 
second is equal to the final velocity minus half of acceleration 
due to gravity. Therefore, this space is found by the formula: 

*■ = v — % g=g v — y 2 ) 

x = Space in feet which the body drops the last second of 

the fall. 
/ = Time in seconds. 
v = Final velocity. 

g = Acceleration of gravity = 32.2 feet. 
h = Height of fall in feet. 

Example. 

A body has in a free fall obtained a final velocity- of 40 feet 
per second. What space did it drop the last second ? 

Solution : 

x —v— Vt g—\^ — — = 40 — 16.1 = 23.9 feet. 

Example. 

A bodv was falling four seconds. How many feet did it 
drop the last second ? 
Solution : 
x = g it — }£) = 32.2 X (4 — %) = 32.2 X 3.5 = 112.7 feet. 



MECHANICS. 279 

TABLE No. 35.— Time, Velocity and Height. 

^=32.161 Feet. 



Time in Seconds. 


Velocity in Feet at 
the End of the Time. 


Height of Fall in 
Feet. 


Distance in Feet that 

the Body Drops in 

the Last Second. 


1 

2 
3 
4 
5 


32.161 
64.322 

96.483 
128.644 
160.805 


16.08 

64.32 
144.72 

257.28 
402.00 


16.08 

48.24 

80.40 

112.56 

144.72 



Upward Motion. 

A body thrown perpendicularly upward with a certain 
velocity will continue the upward movement until it reaches the 
same height from which it would have to fall in order to get a 
final velocity equal to the starting velocity. Therefore, a body 
projected upward with a given velocity will return again with 
the same velocity. This is theoretical in a vacuum, but actually 
the body neither continues to the theoretical height nor returns 
with a final velocity equal to the starting velocity, because the 
air will always offer considerable resistance. The greater the 
weight of a body, in proportion to its volume, the nearer the 
velocity, when it returns, will be equal to its starting velocity. 

Example. 

A body is projected upward with a velocity of 45 feet per 
second. How high will it go before it stops and commences to 
drop again, the resistance of the air not being considered ? 

The solution of this problem is simply to find the theoretical 
height from which a body must drop to attain a final velocity of 
45 feet, which is solved by the formula, 



64.4 



2025 
64.4 



= 31.286 feet. 



Body Projected at an Angle. 

If a body is projected in the direction of the line d e (see 
Fig. 1), with an initial velocity per second equal to the distance 
from d to 1. no force acting after the body is started, it will con- 
tinue to move at constant velocity in a straight line indefinitely ; 
at the end of the first second it would be at 1, at the end of 
two seconds it would be at 2, at the end of the third second at 3, 
at the end of the fourth second at 4, etc.; but, on account of the 
force of gravity, the motion will be entirely different. The 
force of gravity acts on this body exactly as if it was falling in 



280 



MECHANICS. 



a vertical line. At the end of the firsi second the force of 
gravity has caused this moving body to drop 16.1 feet out of its 
path; therefore, instead of being at 1 at the end of the first 
second, it is at a point 16.1 feet vertically under 1 ; instead of 
bsing at 2 at the end of two seconds, it is at a point 2 X 2 X 
16.1 = 64.4 feet vertically below 2; instead of being at 3 at the 
end of the third second, it is at a point 3 X 3 X 16.1 = 144.9 
feet vertically below 3; and instead of being at 4 at the end of 
the fourth second, it is at a point 4 X 4 X 16.1 = 257.6 feet 
vertically below 4, etc. 




When a body is projected in a vertical upward direction 
with an initial velocity of v feet per second, it proceeds to a 



v* 



height -5 — ; therefore, when projected at an angle, a (see Fig. 
* g 

1), with a velocity of v feet per second, it will proceed to the 



height 



v A sin/ a 



2g 



When a body is projected in a vertical upward dirction with 

v 
a velocity of v feet per second, the time for ascent is and the 

time for descent is equal to the time for ascent ; therefore, 

v 
the total time will be 2 ; but when the body is projected 

upward at an angle of a degrees, the total time for ascent and 

2 v sin. a 

descent will be 

g 

The horizontal distance, or the range from d\o n, will be 

equal to the velocity in feet per second multiplied by the total 



MECHANICS. 25 1 

number of seconds consumed in the ascent and descent, and 
this multiplied by cos. of the angle a\ therefore, 

/2 2/ sin. a\ 2 v 2 sin. a cos. a 
Horizontal range = v i J cos. a = 

but 2 X sin. a X cos. a is always equal to sin. of an angle of 
twice as many degrees as the angle a. Therefore, the formula 

, . v 2 sin. 2 a 

^educes to horizontal range = 

g 
Thus, the following formulas will apply to bodies projected 
at an angle. ( See Fig. 1). 

The greatest possible height will be, 

t v 2 sin. 2 a 

The greatest possible rangejwill be, 
7 v 2 sin. 2'a 

g 

The time in seconds will be, 

, 2 v sin. a v sin. a 

g 0.5^ 

v = Velocity in feet per second. 
g=- Acceleration of gravity = 32.2. 

to find the height to which a body can ascend. 

Rule. 

Multiply the velocity in feet per second by the sine of the 
angle (to the horizontal line), square this product and divide 
by 64.4, and the quotient is the height in feet. 

to find the longest possible range. 

Rule. 

Multiply the square of the velocity in feet per second by 
sine of an angle of twice as many degrees as the angle of the 
throw (to the horizontal line), and divide by 32.2. The 
quotient is the longest distance the body can be thrown. 

to find the time of flight. 

Rule. 

Multiply the velocity in feet per second by sine of the 
angle (to the horizontal line), and divide by 16.1. The 
quotient is the time in seconds. 

Example. 

A body is projected at an angle of 55° to the horizontal 
line, with an initial velocity of 120 feet per second. How high 



: r : : H : fir .. :: z: — i : : - :: 1 :i_ i __-t ::: : i : H : 



• : - s . ~ - ; 



" = 



WP X 0L81915 2 



' = 
" = 



1-UflO X O.ffiS 



:: ". : Lz mt »e: :i: :zzi 

= — 



- r : v ..- 






; = 
: — 



1: 1 



jl ::.-.■;: i 1 
A i : : : r 



— . _ 



= = 



mechanics. . 283 

Example 3. 

A nozzle on a hose is placed at an angle of 38° to the 
horizontal line and is spouting water a distance of 40 feet in a 
horizontal direction. What is, theoretically, the velocity of the 
water when leaving the nozzle ? 

Solution : 

v 



=4. 



sin. 2 a 

4 X 32.2 
sin. 76 

v = J40 X 32.2 _ 36 4 feet S econd. 

> 0.9703 

Note. — In Example 2 we multiply by sine of 56 degrees, 

because water is leaving the nozzle at an angle of 28 degrees, 

and twice 28 equals 56. In Example 3 we multiply by sine of 76 

degrees, because twice 38 equals 76. See previous explanations. 

The greatest possible height will be reached if the body is 
thrown perpendicularly upward. The greatest possible range is 
obtained if the body is thrown at an angle of 45° and will then be : 

g 

At an angle of 45° the horizontal range will be twice 
the greatest possible height which could have been reached if 
the body had been thrown perpendicularly upward. At this 
angle the horizontal range is four times the height. For an 
equal number of degrees over or under 45 degrees the horizontal 
range will be equal ; for instance, if a body is thrown out at an 
angle of 30 or 60 degrees, the horizontal distance is the same, 
but the height of ascension will be much more at 60 degrees 
than at 30 degrees. It is frequently useful to notice this in 
practical work. For instance, water under pressure is thrown 
the farthest distance in a horizontal direction from a hose when 
the nozzle is held at an angle of 45 degrees to the horizontal 
line. It is possible by the same pressure to throw water twice 
as far in a horizontal distance as in vertical height. 

Motion Down an Inclined Plane. 

A ball rolling along an incline, as 
a c (Fig. 2), will have the same 
velocity when it gets to c as it would 
have had if dropping freely from a 
to b, supposing all friction to be left 
out of consideration. 

The average velocity will also be 
half of the final velocity, and the 

time used in the fall will be the distance a c (the length of the 
incline), divided by the average velocity per second. 




284 MECHANICS. 

Body Projected in a Horizontal Direction From an 
Elevated Place. 

When a body is projected in a horizontal direction from a 
place which is higher than the one where it strikes the ground, 
the range in feet in a horizontal direction will be equal to the 
product of velocity in feet per second and the time in seconds 
which it will take for a body in a free fall to drop a distance 
equal to the difference in vertical height between the two 
places. Thus : 

Horizontal range = v A/— — 

* g 
v = Initial velocity in feet per second. 

h = Vertical height in feet. 

g= Acceleration of gravity = 32.2 feet. 

Example. 

Water spouts from a nozzle in a horizontal direction at a 
velocity of 30 feet per second and the nozzle is placed 12 feet 
above the ground. What is the horizontal range of the water ? 

Solution : 

Horizontal range = v a/— = 30 a/ 12 X2 = 22.45 feet. 

v g * 32.2 



To Calculate the Speed of a Bursted Fly=Wheel from 
the Distance the Fragments are Thrown. 

The angle of 45 degrees is the one most favorable to the 
range ; therefore, suppose the fragments to leave the wheel at 
that angle and use the formula, 

v 2 _ 

Horizontal distance = b = -— which transposes to v = a,/^ „ 

Rule. 

Multiply the horizontal distance by 32.2, and the square root 
of the product is the slowest possible rim-speed the wheel could 
have had at the time of the accident. 

Example. 

A 30-foot fly-wheel bursts from the stress due to centri- 
fugal force, and fragments were thrown a distance of 300 feet 
from the place of accident. What was the slowest possible 
speed the wheel could have had at the time the accident 
occurred ? and what was the corresponding number of revolu- 
tions per minute ? 



MECHANICS. 285 

Solution : 



V 



ltion : 
V300 X 32.2 = V 9660 = 98.3 feet per second. 



The length of the circumference of a 30-foot wheel is 94.25 
feet, therefore the fly-wheel was running at a speed not less than 

98.3 
60 X q , 25 — 62.6 revolutions per minute. This calculation 

does not prove that the wheel did not run faster than 62.6 
revolutions per minute when it burst; it may have revolved a 
great deal faster, as it is not at all sure that any fragments left 
the wheel at an angle of 45 degrees, but it is certain that the 
speed of the wheel was not slower. Sometimes it may be pos- 
sible to settle upon the angle at which a certain fragment left 
the wheel by noticing traces and marks where it went, and, 
figuring from the angle and the range, a pretty fair idea of the 
bursting speed may be obtained. (See formula on page 283). 



Force, Energy and Power. 

Force is a pressure expressed in a push or a pull. 

Energy is the ability to do work. It is divided into poten- 
tial energy and kinetic energy. 

Poteiitial e?iergy is the ability of a body to perform work 
at any time when it is set free to do so. 

Kinetic energy is the ability of a moving body to do work 
when its motion is arrested. Kinetic energy is very frequently 
called "stored-up energy." 

Work is overcoming resistance through space. In the 
English system of weights and measures the common unit of 
work is the foot-pound. 

Power is the rate of doing work. Work is an expression 
entirely independent of time, but power always takes time into 
consideration. For instance, to lift one pound one foot is one 
foot-pound of work, no matter in what time it is done, but it 
takes 60 times as much power to do it in one second as it would 
^ake to do it in one minute. 

Inertia. 

Inertia is the inability of dead bodies to change either their 
state of rest or motion. In order to bring about any change, 
either of motion or rest, dead bodies must always be acted upon 
by some outside force. 

Resistance due to inertia is the resistance which a dead 
body free to move presents to any external force acting to 
change either its state of motion or rest. 



286 MECHANICS. 



Mass. 

The mass of a body is the quantity of matter which it con- 
tains. By common consent the unit of mass is, in mechanics, 
considered to be that quantity of matter to which one unit of 
force can give one unit of acceleration in one unit of time ; 
therefore, when the weight of a body is divided by acceleration 
of gravity, the quotient is the mass of the body. Thus : 

W 
m = — 

g 

IV = m X g 

W 
g= — 
in 

Momentum. 

The product of the mass of a moving body and its velocity 
is called its momentum or, also, its quantity of motion. The 
unit for momentum is the product when unit of mass is multi- 
plied by unit of velocity per second. In mechanical calcula- 
tions, using English weights and measures, the unit of mass is 
weight divided by 32.2; therefore, unit of momentum will be: 
Weight of the moving body in pounds multiplied by velocity in 
feet per second and the product divided by 82.2. Thus : 

q — m X v m = mass = — ; therefore, 

g 

IV 
q = v 

g 

q=z^L W 

g 
q = Momentum, or quantity of motion. 

W = Weight of moving body in pounds. 

v = Velocity of moving body in feet per second. 

g =■ Acceleration of gravity. 

v 
is the formula by which the time in a free fall is 

8 

obtained, and, consequently, the momentum of a falling body 

can also be expressed by the product of the weight of the body 
in pounds and the time in seconds during the fall. This product 
is usually called " time effect." 

Impulse. 

The product of the force and the time in which it is acting 
as a blow against a body is called impulse, and it is always 
of the same numerical value as the momentum of the moving 
body. 



MECHANICS. 287 

Kinetic Energy. 

The kinetic energy stored in any moving body is always 
expressed in foot-pounds, by the product of the force in pounds 
acting to overcome the inertia of the body, and the distance in 
feet through which the force was acting in starting the body, 
and is always equal to the weight of the body multiplied by the 
square of the velocity and this product divided by twice the 
acceleration of gravity. Thus : 

K == Kinetic energy in foot-pounds. 

W = Weight of the body in pounds. 
v = Velocity of the body in feet per second. 
2 g = 64.4. 
In a free fall the height, h, corresponding to a given 

velocity, is found by the formula, ~x~ ; therefore, K '= W X h. 

Thus, multiplying the weight of a moving body by the height 
which in a free fall corresponds to its velocity, the product will 
be the kinetic energy stored in the body. 

The formula K = — ^ transposes to K = y z ?n v 1 . 

Hence the simple rule : 

Multiply half the mass of a moving body by the square of 
its velocity in feet per second, and the product is the kinetic 
energy in foot-pounds stored in the body. 

The kinetic energy stored in. any moving body always 
represents a corresponding amount of mechanical work which 
is required in order to again bring the body to rest. 

Example. 

A body weighing 1610 pounds is moving at a constant 
velocity of 18 feet per second. How many foot-pounds of 
kinetic energy is stored in the body ? 

Solution : 

„ WXv* 1610 X 18 X 1 8 
K = — Q = A44 = 8,100 foot-pounds. 

If this moving body was brought to rest and all its stored 
energy could be utilized to do work it could lift 8,100 pounds 
one foot, or it could lift 81 pounds 100 feet, or any other combi- 
nation of distance and resistance which, when multiplied by 
one another, will give 8,100 foot-pounds. 

It is very important always to keep in mind a clear dis- 
tinction between work and power, as power is the rate of doing 
work, and time must, therefore, always be considered in the 
question of power. For instance, when 33,000 foot-pounds of 



2 , ; ? MECHANICS. 

work is performed in one minnte it is said to be one horse-power ; 
therefore, if this 32.400 foot-pounds of energy was utilized to do 
work and used up in one minute, it would do work at a rate of 
|| -"— = |i horse-power, but if utilized during a time of two 
minutes it -^ould only do work at a rate of f| horsepower, or if 
utilized in a second die rate of work would be §A X 60 = 58 ^f 
:.::-z-z : .ver. e:z. 

To Calculate the Force of a Blow. 

The force of a blow may be calculated by. the change 
it produces. For instance, a drop-hammer weighing S00 pounds 
drops three feet and compresses the hot iron on the anvil X 
inch. How much is the average force? Qi inch = *4sfoot). 

I he kinetic energy stored in the hammer at the moment it 
: :r:.r:_±~ zes :: zzrzzzzress the iror: is i-Ov ■ : : . = iiOv :: DZ-zozmcis. 
„- " '■■'< 

The average force = — = 115.200 lbs. 

- i 

- * 

In the above example, friction is neglected. 

The shorter the duration of the blow the more intense it 
will be. Therefore the force of the hammer mentioned above , if , 
instead of striking against hot iron, compressing it % inch, had 
been struck against cold iron, compressing it only a few thou- 
sandths, the blow would have been as many times more intense 
as zhe duration of the blow had been shorter. Therefore it is 
entirely meaningless to say that a drop-hammer or any other 
similar machine is giving a blow of any certain number of 
pounds by falling a certain number of feet, because the in- 
tensity of the blow will depend upon its duration. 

Formulas for Force, Acceleration and Motion. 

pound of force acts upon one pound of matter it produces an 
azzelerazizr. •:: V- : - zee: per seczr.z: ezzz'.z szzzzessive sezozz :. as 
long as the force continues to act. 

From Newtons laws of motion, it is known that the motion 
is always in proportion to the force by which it is produced ; 
therefore, when one pound of force acts for one second upon - . 
pounds of matter, it will produce an acceleration of one foot per 
second. 

Hence the following formulas 

m =■ Mass of the moving body, which is considered to be 



weight divided by 






F = Constant force in pounds acting on a body free tc 
move. 

G = Constant azzelerazzorz in z'ee: rer second due to the 
acting force. F. 



MECHANICS. 289 

T = Time in seconds in which the force F acts upon a 
body free to move. 

v — final velocity acquired by the. moving body in the 
time of /^seconds. 



F 

m =■■ — . 

G 


G = Z_ 

in 




v= TG 


~~G 


v . ..'.' F 
T m 




v m = F T 


FT 
m — 


F— v 


m 


7, vm 



F= m G 



T 

FT 

v = 

m v T F 

When a moving body is arrested the product of the resist- 
ance and time is equal to its momentum. Thus: 



R T= v m 


R T 
v — ?n 
m 

7 , v m j. v m 


_RT 

V 



T R 

R = Constant resistance in pounds acting against the 
moving body. 

The average velocity of the moving body is half of the final 
velocity, and the space passed over by the moving body when 
acquiring the given velocity is half of the final velocity in feet 
per second multiplied by the time in seconds. Thus : 

S=-^XT v = 2 -* FJL = *1 

2 T m T 

FT 2 = 2Sm J — \ — -p— -J** 



FT* 
m — 

26* 


S = 


FT 2 

2 m 


S = Space in feet. 







The work in foot-pounds required to overcome the inertia of 
a given body when brought from a state of rest to a given 
velocity is equal to the kinetic energy stored in the moving 
body. Thus : 

jf _ ^ y i>; i>- F v T G m v T 

— - 7) g ~" 2 

K = Kinetic energy stored in the moving body. 



29O MECHANICS. 

The force required to obtain a given velocity in a given 
time, when both resistance due to inertia and resistance due to 
friction is considered, is calculated by the formula : 

Force = f ^? ** X Mass^ + (weight X coefficient of friction). 
V Time / 

which may be written : 

Force = ( — X Mass \ + (resistance due to friction). 

V Time J 

Important. — Always calculate the force required to over- 
come the resistance due to inertia and the force required to 
overcome the resistance due to friction separately, and add the 
two forces in order to obtain the total force required. 

It is sometimes assumed that adding so much to the mass, 
as 3V of the product of weight and coefficient of friction, should 
give the result in one operation : but such an assumption is 
erroneous, because the correct value lor the required force is : 

F _v X W _j_ Jrx/ - 
TXg 

which cannot be transposed to 

F _ v X JF-h Ji'Xf 
TXg 

F = Required force : v = velocity ; T= time ; W= weight 
of moving body in pounds ; g — acceleration due to gravity, or 
S2.2 : /"= coefficient of friction. 

Example. 1. 

A railroad train weighing 225.400 pounds is started from 
rest to a velocity of 50 feet per second : the road is straight and 
level : the resistance due to friction is assumed to remain con- 
stant and to be 1000 pounds. What average constant pull in 
pounds must be exerted by the locomotive at the draw-bar in 
order to bring the train up to this speed in 40 seconds ? 

Solution : 

For the inertia, 

velocitv X mass _ -,-, v 2?5400 

Force = z=— OKJ * 32^ 

fame — = ST50 lbs. 

For friction the force = 1000 

Total force, 9750 lbs. 

Note. — This constant force of 9750 pounds has been acting 
under a uniformly increasing velocity from rest or nothing at 
the start to 50 feet per second at the end of 40 seconds: therefore, 
the average velocity has been half of the hnal velocity, or 25 feet 
.. ,-econd. The average work of the locomotive in starting the 



MECHANICS. 2$f 

train during this 40 seconds was 25 X 0750 = 243,750 foot- 
pounds per second, and the horse-power exerted by the locomo- 
tive on the draw-bar in starting this train was M^W— — 443. IS 
horse-power, but the power required to keep this train in motion 
at a speed of 50 feet per second on a level road will be only 
50_x jooo __ got)} horse-power. From this it may be seen what an 

immense power there has to be produced in order to start heavy 
machinery in a short time, in comparison to the power required 
to keep it going after it is started. 

Example 2. 

How far did the train move before it got up to the required 
speed of 50 feet per second? 

Solution : 

S=^_ = b0X40 = 1000 feet. 

2 2 

Example 3. 

Suppose that after the train had acquired this speed of 50 
feet a second, the locomotive was detached and that the resist- 
ance due to friction continued to be 1000 pounds. How many 
seconds would the train be kept in motion by its momentum on 
a level road ? 

Solution : 

225400 

Time = "^L _ a° X "W 

j? — TqqTj = 3o0 seconds. 

Example 4. 

How many foot-pounds of kinetic energy is stored in this 
train, which weighs 225,400 pounds and runs at a constant speed 
of 50 feet a second ? 

K _ v 2 X m _ 50 2 X -32^- _ 8j 750 )000 foot-pounds. 

Example 5. 

How far will this kinetic energy drive the train on a hori- 
zontal road if we suppose the constant resistance due to friction, 
as in Example 3, to be 1000 pounds ? 

Solution : 
Distance = kinetic energy = 8750000 = S75Q feef 
resistance 1000 

When a body free to move is acted upon by a constant 
force the space passed over increases as the square of time. 



2g: 



MECHANICS. 



Example 0. 

Under the influence of a constant force a body moves five 
feet the first second. How far will it move in eight seconds, 
friction not considered? 

Solution : 

Distance = S 2 Xo = 320 feet. 

Centers. 

Center of gravity is the point in a body about which all its 
parts can be balanced. If a body is supported at its center of 
gravity the whole body will remain at rest under the action of 
gravity. 

Center of gyration is a point in a rotating body at which 
the whole mass could be concentrated (theoretically) without 
altering the resistance, due to the inertia of the body, to angular 
acceleration or retardation. 

Center of oscillation is a point at which, if the whole matter 
of a suspended body was collected, the time of oscillation 
would be the same as it is in the actual form of the body. 

Center of percussion is a point in a body moving about a 
fixed axis at which it may strike an obstacle without communi- 
cating the shock to the axis. 



Moments. 

The measures of tendency to 
produce motion about a fixed point 
or axis, is called moment. The pro- 
duct of the length of a lever and the 
force acting on the end of it. tending 
toswingit around its center, is called 
the moment of force or the statical 
moment, and may be expressed in 

either foot-pounds or inch-pounds. In Fig. 3, the arm is 18 
inches long and the force is 40 pounds; the moment is 18 X 40 
= 720 inch-pounds, or \y 2 X 40 = 00 foot-pounds. 




m 



Levers. 

When a lever is balanced, the distance a, 
multiplied by the weight u>, is always equal to 
the distance b, multiplied by the force F. 
In a bent lever (as Fig. 5) it is not the length 
of the. lever but the distance from the fulcrum 
at right angles to the line in which the force 
acting, that must be multiplied. Thus: 
a X w = b X F. 

In Fig. 6, the force is acting at a right angle 
to the lever, and, therefore, the distance a is equal 
to the length of the long end of the lever. 



Fig, 



is 



Fig. 5. 

«-a+ — 6— 



M 



MECHANICS. £93 




The force is applied to more advantage in 
Fig. G than in Fig. 5. As a rule, the force should 
always be applied so as to act at right angles to 
the lever. 



Radius of Gyration. 

The radius of gyration of a rotating body is the distance, 
from its center of rotation to its center of gyration. 

„ j. £ ,. A moment of rotation 

Radius of gyration =: -yl s : — _ 

'mass oi rotating body 

or, for a plane surface: 

r, -,. r ,• /moment of inertia 

Radius of gyration = -v/ — 

* area of surface 

The radius of gyration of a round, solid disc, such as a grind- 
stone, when rotating on its shaft, is equal to its geometrical 
radius multiplied by \/>^~or radius multiplied by 0.7071 very 
nearly. The radius of gyration of a round disc, if indefinitely 
thin and rotating about one of its diameters, is equal to radius 
divided by 2. The radius of gyration of a ring, of uniform cross- 
section, rotating about its center, such as a rim of a fly-wheel 
when rotating on its shaft, is : 



— — — 



7? = Outside radius. 
r== Inside radius. 

The radius of gyration of a hollow circle when rotating 
about one of its diameters is : 



\j?2 _i_ r i 
Radius of gyration = -v — — 



7l = Outside radius. 
r = Inside radius. 

Moment of Inertia. 

The moment of inertia is a mathematical expression used 
in mechanical calculations. It is an expression causing con- 
siderable ambiguity, as it is not always used to mean the same 
thing. 

The least rectangular moment of inertia, as used when 
calculating transverse strength of beams, columns, etc., is the 
sum of the products of all the elementary areas of cross-sections 
into the square of their distances from the axis of rotation. 
The axis of rotation is considered to pass through the center 
of gravity of the section. 



294 



MECHANICS. 



The least rectangular moment of inertia is always equal 
to the area of surface of cross-section, multiplied by the square 
of the radius of gyration, when the surface is assumed to rotate 
about the neutral axis of the section. 

Mathematicians calculate the moment of inertia by means 
of the higher mathematics, but it may also be calculated 
approximately by dividing the cross-section of the beam into 
any convenient number of small strips and multiplying the area 
of each strip by the square of its distance from its center-line to 
the neutral axis, and the sum of these products is the moment 
of inertia, very nearly. 

The narrower each strip is taken, the more exact the result 
will be ; but it will always be a trifle too small. 

Example 1. 

Find approximately the rectangular moment of inertia for 
a surface (or section of |a beam) 6" X 2", about its axis x y. 
(See Fig 7.) m 

Divide the surface into narrow strips, as a, o, c, a, e, f, g, 
Aj i,j\ k, /, and multiply each strip by the square of its distance 
from the neutral axis, xy, and the sum of these products is the 
moment of inertia of the surface. 



i X (2X) 2 = 5.0625 
i X (IX) 2 =3.0625 



tf = 2X^X (2%)* = 7.5625 
b = 2 X 
c = 2 X 

d=2 X y 2 X (1%) 2 = 1.5625 

X ( X) 2 = 0.5625 

X ( X) 2 = 0.0625 

X ( X) 2 = 0.0625 

y 2 X ( %)* = 0.5625 

y 2 X (IX) 2 = 1.5625 

y 2 x (IX) 2 = 3.0625 

X (2X) 2 = 5.0625 

l—2Xy 2 X (2X) 2 = 7.5625 

Moment of inertia = 35.75 (approximately). 

The correct value for the least rectangular 



e — 


-2 X 


f= 


--2 X 


g- 


-2 X 


h = 


■ 2 X 


i — 


2 X 


j- 


2 X 


k = 


2 X 




moment of 



inertia for such a surface is obtained by the formula, 



(Depth) 3 X width 



and for Fig. 7 will be 



6 3 X 2 



= 36. Thus, the 



12 ~ 12 

approximate rule gives results a trifle too small, but if the sur- 
face had been divided into smaller strips, the result would have 
been more correct. 

Radius of gyration for this surface, when rotating about 
the axis x y, is : 



V 



moment of inertia 
area 



_ /36 ,— 

- Til" = V 3 = 



1.73 inches. 



MECHANICS. 



295 



Example 2. 

Find by approximation the rectangular moment of inertia 
for a surface, as Fig. 8, (the sectional area of an I beam) about 
the axis x y. 

When the beam is symmetrical, the neutral axis is at an 
equal distance from the upper and lower side, and the moment 
of inertia for the upper and lower half of the beam is equal ; 
consequently, when calculating momentof inertia for a surface 
like Figs. 8 and 7, it is only nec- 
essary to calculate the moment 
of inertia for half the beam, 
and multiply by 2 in order to 
get the moment of the whole 
beam. 



Solution: 



a = 3 
b =3 
c —\ 

e = 1 
/=1 



i X(2X) 2 = 
i X(l#) 2 = 
i X(1X) 2 = 

i x( H) 2 = 
.*x< X) 2 = 

Moment of inertia = 
Moment of inertia = 
Moment of inertia = 43.125 



X 
X 

X 
X 
X 
X 



11.34375 
7.59375 
1.53125 
0.7S125 
0.28125 
0.03125 




21.5625 
21.5625 



4 3.125 
10 



= 2.07 inches. 



for upper half, 
for lower half, 
for beam (approximately). 
Area of cross-section of beam is 10 square inches. 

Radius of gyration = -v/J 

Example 3. 

Find approximately the moment of inertia of a surface, as 
Fig. 9 (usual section for cast-iron beams), about the axis, x y 9 
passing through the center of gravity of the surface. 

In shapes of this kind the axis through the center of gravity 
is not at an equal distance from the upper and lower side, but it 
can be obtained experimentally by cutting a templet to the 
exact shape and size of the surface and balancing it over a 
knife's edge, or it maybe calculated by the principle of moments, 
as shown in this example. Divide the surfaces into three 
rectangles, the upper flange, the web and the lower flange. 
Assume some line as the axis, for instance, the line n m, which 
is the center line through the lower flange; multiply the area of 
each rectangle by the distance of its center of gravity from the 
axis ?i m, and add the products. Divide this sum by the area of 
the entire section, and the quotient is the distance between the 
center of gravity of the section and the axis n m. 



296 



MECHANICS. 



* 2- 



FIG. 9. 



i 7Z- 



1" *j 38 

5 3C_ V 







SOLVING FOR CENTER OF GRAVITY : 

(Area.) (Distance.) 

Area of upper flange = 2 XI = 2 square inches X 5 =10 
Area of web = 4 XI = 4 square inches X 2 l 2 = 10 

Area of lower flange =4X1 = 4 square inches X = 

10 20 

and 20 divided by 10 = 2" which is the distance from the center 
of gravity of the lower flange to center of gravity of the section 
of the beam, or the neutral axis x y. 

SOLVING FOR .MOMENT OF INERTIA : 

a=2 X ) 2 X (3; 4 j 2 = 10.56250 
b — 2X} 2 X (;2Uy 2 = 7.56250 
c = l X ) 2 X (2U) 2 = 2.53175 
d=l X y 2 X (IH)' 2 = 1-53225 
e = l X / 2 X (IX)' 2 = 0.78125 
f—\ X J 2 X ( U!'= 0.28125 
g=l x % X ( X)-= 0.03125 
* = 1X l / 2 X ( U:-= 0.03125 
f = lX Yz X ( U: 2 = 0.25125 
/=1-X J 2 X (1J0 9 = 0.78125 
£ = 4 X ^ X (l.Vp = 6.12500 
I— 4 X % X h>&= 10.12500 

Moment of inertia of beam = 40.6266 (approximately). 
Area of cross-section of beam = 10 square inches 



Radius of gyration of beam = a 



40.(»2(Hj 

— 10 



= 2.015 inches. 




MECHANICS. 297 

Polar Moment of Inertia. 

The polar moment of inertia is a mathematical expression, 
used especially when calculating the torsional strength of 
beams, shafting, etc. It is very frequently denoted by the letter 
/. The polar moment of inertia is the sum of the products of 
each elementary area of the sur- ^ F|G 10 _ 

face multiplied by the square of 
its distance from the center of 
gravity of surface. Suppose (in 
Fig. 10) that the area is divided 
into circular rings, as a, b, c, d, e, 
f, g, h, h j\ k, I, m, 11, o,p, and the 
area of each ring multiplied by 
the square of its distance from 
the center, c ; the sum of all these 
products is the polar moment of 
inertia. The moment calculated 
this way, will always be a trifle 
too small, but the smaller each 
ring is taken the more correct the 
result will be. If each ring could 
be taken infinitely small the result would be correct. 

The polar moment of inertia is equal to the square of the 
radius of gyration about the geometrical center of the shaft, mul- 
tiplied by the area of cross-section of the shaft; therefore, for a 
round, solid shaft (as the section shown in Fig. 10), the polar 
moment of inertia is always expressed by the formula : 
(Radius) 4 X ^ Qr (Diameter) 4 X * 
2 ' 32 

For a hollow, round shaft, the polar moment of inertia is 
expressed by the formula, 

(Z*-</* ) y 

J 32 

D = Outside diameter. d = Inside diameter. 

The fundamental principle for the polar moment of inertia 
for any shape of section is that, if two rectangular moments of 
inertia are taken, one being the least rectangular moment of 
inertia, about an axis passing through the center of gravity, and 
the other, the least rectangular moment, about an axis perpendic- 
ular to the first one, also through the center of gravity, the sum 
of those two rectangular moments is equal to the polar moment. 

In Fig. 10, the rectangular moment of inertia about the axis 

x y will be ( meter ) 4 X ^ and the rectangular moment about 
J 64 

the axis x f y' will also be ^ ia m _g g r ) x " ; thus the polar moment 

will be (^^eter) 4 _X_rr 
32 



2<?8 mechanics. 

Example. 

Find the polar moment of inertia and radius of gyration of a 
round shaft of 4" diameter. 



Solution 



J= D^ 



32 

32 



/= 4*X 3.1418 = 251328 



Radius of gyration = J Pokr moment of inertia 



25.1328 



area of section. 

Radius of gyration = \L 

*2 2 X 3.1410 

Radius of gyration = 1.414 inches. 

The term, moment of inertia, as used in calculating stored 
energy in revolving bodies, is frequently and certainly more 
concisely called moment of rotation, and is a mathematical 
expression by which the effect of the whole mass (theoretically) 
is transferred to the unit distance from center of rotation. This 
term (moment of inertia or moment of rotation) is obtained by 
multiplying the square of radius of gyration by mass of moving 
body.* In English measure, mass is taken as * ^ of the weight 

of the revolving body, and the radius of gyration is always taken 
in feet. 

Example. 

A solid disc of cast-iron, rotating about its geometrical 
center, is six feet in diameter and of such thickness that it will 
weigh 4073.3 pounds. What is its moment of rotation or 
moment of inertia? 

Radius of gyration = 3 xV} and (radius of gyration) 2 = 3 2 X \ 

Mass= 4078 - 3 = 126.5 

32.2 

Moment of rotation = 126.5 X 3 2 X y 2 = 569.25. 

Note. — In all such problems relating to stored energy in 
rotating bodies, the radius of gyration is usually taken in feet 
and not in inches, as in previous examples of moment of inertia, 
when relating to strength of material. 

* Instead of multiplying the mass of the body by the square of radius of 
gyration in feet and calling the product moment of inertia, some writers multiply the 
weight of the body by the square of the radius of gyration in feet and call this 
product moment of inertia. This last expression for moment of inertia, of course, 
will have a numerical value of '32.2 times the first one. It does not make any differ- 
ence in the result of the calculation whether weight or mass is used, but the same 
unit must be adhered to throughout the whole calculation. 



MECHANICS. 299 

Angular Velocity. 

When a body revolves about any axis, the parts furthest 
from the axis of rotation move the fastest. The linear velocity 
at a radius of one foot front the center of rotation is called the 
angtdar velocity of the body. It is usually reckoned in feet per 
second. The angular velocity of any revolving body is ex- 
pressed by the formula, 

V* = 2 it n 

V % = Angular velocity in feet per second. 
n = Number of revolutions per second. 

Rule, 

Multiply the number of revolutions per second by 6.2832, 
and the product is the angular velocity in feet per second. 

Example. 

What is the angular velocity of a fly-wheel making 300 
revolutions per minute ? 
Solution : 

300 revolutions per minute = 5 revolutions per second, 
therefore, angular velocity = 6.2832 X 5 = 31.416 feet per 
second. Angular velocity expresses the velocity at unit dis- 
tance from center of rotation and in English measure this unit is 
feet. As already stated, the moment of rotation is an expres- 
sion for the mass of the rotating body (theoretically) transferred 
to unit distance from center of rotation ; the product of angular 
velocity and moment of rotation will, therefore, be the 
momentum of the rotating body. The constant resistance 
which has to be exerted at unit radius in order to bring the 
body to rest in T seconds will be : 

T 

The resistance which has to be exerted at any radius of r 
feet to bring the body to rest in T seconds will be : 

r T 
R = Resistance in pounds. 
Va, = Angular velocity in feet per second. 
/ = Moment of rotation (also called moment of inertia). 

The constant force which has to be exerted at unit radius 
in order to bring the body from a state of rest to an angular 
velocity V* in T seconds will be : 

zr_ K I 



300 :z ;:-:amcs. 

The constant force which has to be exerted at any radius, 
r. in order to bring the body from a state of rest to an m rilar 
velocity V % in T seconds wiH t k 

i? = Constant resistance in pc 7- : = 

F = Constant force in pounds. 

V m = Angular velocity in feet per seconcL 

/ = Moment of rotation (also called moment of inertia). 

r = Radius in feet at which the force is applie : 
T = Time in seconds that the for: t is : : zz. z 

Ftamp if 

A -..- ■-- ::i making 120 revolutions per minute and weigh- 
ing 433 pounds, is brought to rest in tw a 2 : : 3y a resistance 
acting a: i - :-:7_ch radius. The radius of 7-717:7 :: the ~; • 
whed I . :-■ What is the average force exerted against the 
resis::::i :t luring these twc secon 

- : . :~ . : :. 

:--- . ": Dns per minute = 2 revolutions per second. 

.-. t-gular veloc : tj = . : . . = 13 ' Xi feet ter second. 

rn: :: :;-::;_: = l._ L . jr -° =il : 

_ 

Radius of resistance. 6 inches = 0.5 feel 

R='z=l: : Ji-^2- =--. rounds. 

: 

If a rotatin r s not brought to rest, but only reduced 

in speed to an angular veloc itj :: ~\ a _~~ sec ands then the 
aven r - - : : : : - m res s Ean : - act ng at unit radius is 



The average force which has to be taunted at any radius at 

: _et to reduce the angular vel : -::~ : : ~\ in 7 = e : jnds will be : 

- _ (t\ — l\j) I 

Example. 

-_ ~ - heel on a punching machine weighs 644 pounds, its 

:> f _— t id :n is 1 L - feet, md m : •: e ; it normal speed §00 

- - ifcntions per minute, but when, the machine is c -inching the 



MECHANICS. 30I 

speed is in A of a second reduced to a rate of 280 revolutions 
per minute. What average force has the fly-wheel communi- 
cated to the pitch-line of a 6-inch gear on the fly-wheel shaft ? 

Solution : 

The mass of the fly-wheel = — — = 20 

32.2 

The moment of rotation = (l^) 2 X 20 = 45 

300 revolutions per minute = 5 revolutions per second. 
Angular velocity = 5X 6.2832 = 31.416 

280 revolutions per minute = 4% revolutions per second. 

Corresponding angular velocity = 4^3 X 6.2832 = 29.3216 

6-inch diameter of gear = 3-inch radius = % foot. 
F _ (31.416 — 29.3216) X 45 

*X % 
F = 2.0944 X 45 X 5 X 4 = 1884.96 pounds. 

The kinetic energy in foot-pounds stored in the revolving 
body may be obtained by the formula : 

Fa 2 X — = kinetic energy. 

Decreasing the angular velocity to Vai, the stored-up 
energy will also decrease to 

Fal 2 X -L 

2 

and the work done by the revolving body will be 

/ 



(Fa 2 - Fal 2 ) X 

2 



Example 1. 



The moment of rotation in a fly-wheel is 1040 ; its angular 
velocity is 5 feet per second. What is the stored-up energy in 
the wheel ? 

Solution : 

Kinetic energy = 5 2 X 1040 = 13,000 foot-pounds. 

Example 2. 

At certain intervals, when machinery is started, the angular 
velocity of this fly-wheel is reduced to 4J^ feet per second. 
How many foot-pounds of energy has the fly-wheel given up in 
helping to drive the machinery ? 



302 MECHANICS. 

Solution : 

* = (5* - (4^)2) X 520 

x — (25 — 20 X) X 520 

x = 4% X 520 = 2470 foot-pounds of energy given 
out by the fly-wheel during this change of speed. 

Example 3. 

How much stored energy is left in the wheel after its angu- 
lar velocity is reduced to \y 2 feet per second ? 

Solution: 

K= (±y 2 Y 2 X 520 = 20 % X 520 = 10,530 foot-pounds. 
The same result may be obtained by subtracting, thus : 
13.000 — 2470 = 10.530 foot-pounds. 

Centrifugal Force. 

The centrifugal force is the force with which a revolving 
body tends to depart from its center of motion and fly in a 
direction tangent to the path which it describes. The centrip- 
etal force is the force by which a revolving body is prevented 
from departing from the center of motion. When the centri- 
fugal force exceeds the centripetal force the body will move 
away from the center of motion, but if the centripetal force ex- 
ceeds the centrifugal force, the bod}- will move toward the center 
of motion. The centrifugal force in any revolving body is equal 
to the mass of the body (see page 2$6) multiplied by the square 
of its velocity, and this product divided by the radius of the 
revolving body. 

cf— W X v ~ — m X v2 
~ 32.2 X r — r 

cf= Centrifugal force in pounds. 

r = Radius in feet. 

v = Velocity in feet per second. 
W = Weight of moving body in pounds. 
m = Mass of moving body. 

Example. 

The weight a, in Fig. 11, is four pounds, and the length of 
the string is two feet : the weight is made to swing around the 
center c, three revolutions per second. What is the stress on 
the string due to centrifugal force ? 




MECHANICS. 303 

Solution : 

The distance from c, to the center of the ball is two feet, 
and making three revolutions per second, the velocity will be 
2 X 3 X 3.1416 X 2 = 37.7 feet per second. 

tf= 4X 37 - T X 37 - 7 = SS.2 pounds. 
32.2 X 2 

In metric measure, 

, W X v* 
c f— 

9.81 X r 

cf= Centrifugal force in kilograms. 

r = Radius in meters. 

v =. Velocity in meters per second. 
W=- Weight of moving body in kilograms. 

Example. 

Suppose that the weight a. in Fig. 11. is live kilograms, 
swinging around the center, c, one revolution per second ; the 
distance from a to c is \% meters. What is the stress on the 
string due to centrifugal force ? 

Solution : 

The velocity will be 1.5 X 3.1416 X2 = 9.4248 meters per 
second. 

5 X 9.4248 X 9.4248 = ^ kilograms . 

9.si x iy 2 



Friction. 

The resistance which a body meets with from the surface 
on which it moves is called friction. It is called sliding friction 
when one body slides on another: for instance, a sleigh is pulled 
along on ice — the friction between the runners of the sleigh and 
the ice is sliding friction. It is said to be rolling friction when 
one body is rolling on another so that new surfaces continually 
are coming into contact; for instance, when a wagon is pulled 
along a road, the friction between the wheels and the road is roll- 
ing friction, but the friction between the wheels and their axles 
is sliding friction. Sliding friction varies greatly between 
different materials, as everybody knows from daily observation. 
For instance, a sleigh with iron runners can be pulled with less 
effort on ice than on sand, even if the road is ever so smooth. 
This is because the friction between iron and ice is a great 
deal less than the friction between iron and sand, 






300 lbs. 



304 MECHANICS. 

Coefficient of Friction. 

The ratio between the force required to overcome the re- 
sistance due to friction and the weight of a body sliding along 
a horizontal plane is called coefficient of friction. 

For instance, in Fig. 12 a fig. 12 

piece of iron weighing 300 lbs. 
rests on a horizontal plate b. A 
string fastened to a, goes over a 
pulley, c. At the end of the 
string is applied a weight, d. If ^ a 
this weight is increased until r^\ - Q ^ 

the body a just starts to move * 

along on b, and the weight is found to be 50 pounds, the co- 
efficient of friction will be - 50 — — = 0.166 

300 — 6 

When the weight of a moving body is multiplied by the 
coefficient of friction, the product is the force required to keep 
the body in motion. Of course, any pressure applied to the 
moving body, perpendicular to its line of motion, may be sub- 
stituted for its weight. For instance, the frictional resistance 
of the slide in a slide-valve engine is not due to the weight of the 
valve, but to the unbalanced steam pressure on the valve. In all 
cases the rule is : 

Multiply the coefficient of friction by the pressure perpen- 
dicular to the line of motion, and the product is the force 
required to overcome the frictional resistance. 

Example. 

The coefficient of friction is 0.1, and the weight of the 
sliding body is 800 pounds. What force is required to slide it 
along a horizontal surface ? 

Solution : 

Force = S00 X 0.1 = 80 pounds. 

Rolling Friction. 

If the body, a, (see Fig. 12) was lifted up from the plane, b, 
high enough so that two rollers could be placed between a and 
b, it would be found that the body would move with much less 
force than 50 pounds because, instead of sliding friction, as in 
the first experiment, it would be rolling friction. Suppose it is 
found that a commenced to move when the load, d, was four 
pounds, then the coefficient of friction for this particular case 

would be — = — = 0.0133 
300 75 

In these experiments the whole force at d is not used to 

move the load a, as a small part of it is used to move the pulley 

at c, but in order to make the principle plain, this loss has not 

been considered. 



MECHANICS. 305 

Axle Friction. 

The friction between bearings and shafts is frequently 
called axle friction. This, of course, is sliding friction, but 
owing to the fact that the surfaces in question are usually very 
smooth and well lubricated, the coefficient of friction is smaller 
than for ordinary slides. 

Example 2. 

A fly-wheel weighs 24,000 pounds, the diameter of the shaft 
is 10 inches, and the coefficient of friction in the bearings is 0.08. 
What force must be applied 20 inches from the center in order to 
keep the wheel turning ? 

Resistance due to friction = 24000 X 0.08 = 1920 pounds. 

This resistance is acting at a radius of 5 inches, but the 
force is acting at a radius of 20 inches ; therefore, the required 
force necessary to overcome friction will be x 5 = 480 pounds. 

How much power is absorbed by this frictional resistance if 
the wheel is moving 72 revolutions per minute ? 

Solution : 

The space moved through by the force is 72 x 20 x 2 x 3.1416 

= 753.984 feet, and 753.934 X 480 = 361,912.32 foot-pounds and 
361912^32 _ 10 97 horse-power. 

33000 r 



Horse=Power Absorbed by Friction in Bearings. 

The horse-power absorbed by the friction in the bearings 
for any shaft may be figured directly by the formula, 

Hp _ W XfX n X 3.1416 X d 

•33000 X 12 

This reduces to : 

H-P — WXfXnXdX 0.000008 
H-P = Horse-power absorbed by friction. 
W = Load on bearings in pounds. 
d= Diameter of shaft in inches. 
f= Coefficient of friction. 
n = Number of revolutions per minute. 

Calculating the previous example by this formula, we have: 

H-P = 24000 X 0.08 X 72 X 10 X 0.000008 = 11.06 horse- 
power, which is practically the same as figured before. 



306 MECHANICS. 

Angle of Friction. 

Suppose, instead of using the string and the weight d ''see 
Fig. 12), that one end of the plane is lifted until a commences to 
slide : the angle between b and the horizontal line, when a com- 
mences to move, is called the angle of friction. The coefficient 
of friction may also be calculated from the angle of friction, thus : 
If the body commences to slide under an angle oia degrees, the 
coefficient of friction will be sm " a = tang. a. Thus, the coefficient 

cos. a 

of friction is always equal to tangent of the angle of friction. 

Rules for Friction. 

1. Friction is in direct proportion to the pressure with 
which the bodies are bearing against each other. 

2. Friction is dependent upon the qualities of the surfaces 
of contact. 

3. The velocity has. within ordinary limits, no influence on 
the value of the coefficient of friction. 

■A. Sliding friction is greater than rolling friction. 

5. Friction offers greater resistance against starting a body 
than it does after it is set in motion. 

6. The area of surfaces of contact has. within ordinary 
limits, no influence upon the value of the coefficient of friction. 
but if they are unproportionally large or small the friction will 
increase. 

TABLE No. 36. — Coefficient of Friction. 





SU"E5. 


Bearings. 


Materials. 


Well Not well 
Lubri- Lubri- 
cated, cated. 


Well Not well 
Lubri- Lubri- 


Cast-iron on wrought iron .... 

Cast-iron on cast-iron 

Wrought iron on brass 

Wrought iron on wrought iron . . 


0.08 0.16 
0.08 0.16 
0.08 0.20 
0.10 0.20 


0.05 0.075 
0.05 0.075 
0.05 0.075 
0.05 0.075 



Friction in Machinery. 

When the surfaces are good the frictional resistance for 
slides may be assumed as 10 per cent., more or less, according 
to the conditions of the surfaces. It is always well not to take 
the coefficient of friction too small: it is better to be on the 
safe side and allow power enough for friction. In bearings for 
machinery, the frictional resistance ought not to absorb over 
six per cent. If more is wasted in friction, there is a chance 
for improvement. 



MECHANICS. 



307 




150 lbs. 



150 lbs. 



Pulley Blocks. 

When friction is not considered, the fig. 13. 

force and the load will be equal in a single 
fixed pulley (as A, Fig. 13). 

Thus, a single fixed pulley does not 
accomplish anything further than to change 
the direction of motion. In a single movable 
pulley (as ; at B, Fig. 13), the force is equal 
to only half the load; thus, 75 pounds of 
force will lift 150 pounds of load, but the 
force must act through twice the space that 
the load is moved. The tension in any part 
of the rope in B is half of the load IV; thus, when the load is 
150 pounds the tension in the rope is T5 pounds, when arranged 
at B, but it is 150 pounds when arranged as at A. 

Fig. 14 shows a pair of single sheave pulley blocks in 
position to pull a car; when the blocks are arranged as at A, 
and friction is not consid- 
ered, a force of 100 pounds 
on the hauling part of the 
rope exerts a force of 300 
pounds on the post but only 
200 pounds on the car; but. 
turning the blocks end for 
end, as shown at B. a force 
of 100 pounds on the hauling 
part of the rope exerts a 
force of 300 pounds on the car and 200 pounds on the post. 
This is a point well worth remembering when using pulley blocks. 
Suppose, for instance, that a man exerted a force of 100 
pounds on the hauling part, and that it required 250 pounds 
of force to move the car ; if he used the pulley blocks as 
shown at A, his work would be useless, as far as moving the car 
is concerned, as he could not do it, but turning his blocks end 
for end he could accomplish the desired result. Always re- 
member whenever it is possible to have the hauling part of the 
rope coming from the movable block and pull in the same 
direction as the load is moving. 




Friction in Pulley Blocks. 

In practical work, friction will have some influence, and, to 
a certain extent, change these results, because some of the ten- 
sion in the rope is lost by friction in each sheave the rope passes 
over, therefore the tension in each following part of the rope is 
always less than it was in the preceding part. This loss must 
be obtained from experiments. In good pulley blocks, having 
roller bearings, this loss is probably not more than 0.1, and we 



;o3 



MECHANICS. 



get a useful effect of 0.9 of the force from one part of the rope 
to the next ; therefore, when friction is considered, the useful 
erf ect in the following cases will be : 

In single sheave blocks having the hauling part from the 
movable block (pulling with the load as in B. Fig. 14). 

\V—F{\ 4 0.9 + 0.9 2 ) 
IV =F X 2.71 

In single sheave blocks having the hauling part from the 
fixed block (pulling against the load as in A. Fig. 14), 

IV = F (0.9 + 0.9 2 ) 
IV=FX 1.71 

In double sheave blocks having the hauling part from the 
movable block. 

W = F (1 4- 0.9 + 0.9 2 4 0.9 3 4 0.9 4 ) 
IV =F X 4.1 
In double sheave blocks having the hauling part from the 
fixed block. 

W—F (0.9 4 0.9 2 4 0.9 3 4 0.9 4 ) 
IV = FX 3.1 



Differential Pulley Blocks. 

In a differential pulley block (see Fig. 15). the 
proportion between the force and the weight. 
when friction is neglected, is expressed by the 
formula : 



Fig. 15. 



F = 



IV X (F — r) 
2 X R 



The actual force required to lift a weight by 
such a pulley block is about three times the 
theoretical force, as calculated above. 




Inclined Plane. 

When a weight is pulled upward 
on an inclined plane, as shown in Fig. 
16, and the force F is acting parallel 
to the plane, the required force for 
moving the body will be F = IV X 
sin. a plus friction, and the perpen- 
dicular pressure F, against the plane 
will be IV X cos. a. 



Fig. 16. 




— Cos .-a H 



mechanics. 309 

Example 1. 

The weight, W, (Fig. 16) is 100 pounds ; the angle a is 30°. 
What, force, P, is required to sustain this weight, friction not 
considered ? 

Solution : 

Sin. 30° = 0.5 

Thus : 

p= JV X sin. 30° = 100 X 0.5 = 50 pounds. 

Example 2. 

What is the perpendicular pressure under conditions stated 
in Example 1 ? 

Solution : 

P =W X cos. a = 100 X 0.86603 = 86.6 pounds. 

Therefore, the frictional resistance between the sliding 
body and the inclined plane will be only what is due to 86.6 
pounds pressure ; in other words, the force required to over- 
come friction will be W X f X cos. a. 

Example 3. 

What force is required to move the body mentioned in 
Example 1 when friction is also considered, taking coefficient 
of friction, p, as 0.15? 

Solution : 

P = W (sin. a + cos. a X f) 

F = IW X (0.5 -f 0.86603X0. 15) =100 X 0.6290 =62.99 pounds. 

Note. — This is the force required for moving the load. 
In order to put it in motion more force must be applied, varying 
according to velocity, but after motion is commenced the speed 
would be, under these conditions, maintained forever by this 
force of 62.99 pounds. 

When a load is moving down an inclined plane the force 
due to W X sin. a will assist in moving the body, and if the 
product W X sin. a exceeds the product W X cos. a X f the 
body will slide by itself. For instance, in the body mentioned 
in the previous example, the force required to overcome gravity, 
regardless of friction, is 50 pounds, and the force required to 
overcome friction is 12.99 pounds ; thus, if the body should be 
let down the plane instead of pulled up, it would have to be 
held back with a force of 50 — 12.99 = 37.01 pounds. 

Note. — When the incline is less than 1 in 35, cosine is so 
nearly equal to 1 that it may be neglected, and the force required 
to overcome friction may be considered to be the same as on a 
level plane. For instance, a horse is pulling a load and ascend- 
ing a gradient of 1 in 35 ; if the tractive force required to pull 
the load on a level road was 30 pounds and the weight of the 
load was 1400 pounds, when ascending the hill, the horse will first 



310 MECHANICS. 

have to exert a force of 30 pounds, which is all due to friction, 
but beside that he must also exert a force of T V times 1400 = 40 
pounds : thus the total pull exerted by the horse will be 70 
pounds. 

Inclined Plane With the Force Acting Parallel to the Base. 

When the pressure is continually fig. it. 

acting in a line parallel to the base of 
the incline, as F (see Fig. IT) which 
is frequently the case in mechanical 
movements, as for instance, in screws, 
some kinds of cam motions, etc., it 
will require more force to move the 
body than it would if the force was 
acting parallel to the incline. When 

force acts parallel to the base, as in Fig. 17. the force required 
to move the body, if friction is not considered, will be : 

F= ^Xsin. g = W X tang, a 
cos. a 

Example 1. 
What force is required to move 100 pounds upward an 
incline of 30 c . as in Example 1. excepting that the force is acting 
parallel to the base instead of parallel to the incline ? 
Solution : 

F— W X tang. 30° 

F= W X 100 X 0.57735 = 57.74 pounds. 

When both the friction and the weight of the body are con- 
sidered, the force required to move the body will be : 




F= W X 



sin.tf + (/* X cos. a) 



cos.a — (/X sin.rt) 

Example 2. 

What force is required to move 100 pounds upward an in- 
cline of 30 3 (as in Example 1) if the force is acting parallel to 
the base line instead of parallel to the incline : coefficient of 
friction is supposed to be 0.15? 

Solution : 

F = 100 X sin - 3 ° r + (°- 15 X C0S - 3QC ) 
cos. 30° — (0.15 X sin. 30°) 

^_ 100 x 0.5 + (0-15 X 0.86603) 
0.86603 — (0.15 X 0.5) 

^=ioox - 5 + - 1277045 

0.86603 — 0.075 
F = 100 X 0.7936 = 79.36 pounds. 



MECHANICS. 



311 



Note. — From these calculations it is seen that it is more 
advantageous to apply the force parallel to the incline than 
parallel to the base. When force is applied parallel to the in- 
cline : 

The force required to overcome gravity = 50 pounds. 
The force required to overcome friction = 12.99 pounds. 

Total force = 62.99 pounds. 

When the force is acting parallel to the base : 

The force required to overcome gravity = 57.74 pounds. 
The force required to overcome friction = 21.62 pounds. 

Total force = 79.36 pounds. 

Screws. 

When friction is not considered, the force which may be 
exerted by a screw (see Fig. IS) will be : 

F X R X2~ ^ WX P 



W 



F 



P R X 2 - 

W = Weight. of the load lifted, or force exerted, if the 

screw acts as a press. 
F = Acting force. 

R = Radius in inches at which the force acts. 
P = Pitch of screw in inches. 

FIG. 18 

Regarding friction in 
screws, the thread of a screw 
may be considered as an in- 
clined plane, of which the 
cos. is the middle circum- 
ference of the screw, the 
sin. is the pitch, and the 
force is acting parallel to 
the base. Hence the fol- 
lowing formula : 




F= W X 



P+fXd* 



X 



r 

~R 



drr-fXP 
F = Force, acting at a radius of R inches. 
W — Weight. 

P = Pitch of screw in inches. 
f = Coefficient of friction, usually taken as 0.15. 
R = Radius in inches at which the force is acting. 
r = Middle radius of screw in inches. 
d = Middle diameter of screw in inches. 



3*2 MECHANICS. 

Example. 

Find the force required to act on a lever 30 inches long (see 
Fig. 18) in order to lift the load IV, which is 8000 pounds ? The 
screw is >^-inch pitch and 1^-inch middle radius ; coefficient of 
friction, 0.15. 
Solution : 

F = 8000 X °- 5 + 0-15 X 3.1416 X 2.5 1.25 
2.5 X 3.1416 — 0.15 X 0.5 X 30 

7^=8000 X 1 ; 6781 X 0.0416 = 89.6 pounds. 
7.779 F 

When the screw has V thread, the frictional resistance will 
be increased as -^- of the angle a (see Fig. 18), or equal to 
secant of half the angle of the thread. For United States 
standard screws the angle of thread is 60°, half the angle is 30°, 
and secant of 30° is 1.1547, and the formula will, for United 
States standard thread, become : 

F= JVX P + <t*\.Vbf r 
dir — lAb/P R 

All the letters having the same meaning as in the formulas 
for the square-threaded screws. 

The following table is calculated for square-threaded screws, 
the pitch of the screw being double that of the United States 
standard screw of same diameter. The depth of the thread is 
equal to its width. We see no good reason why the depth of a 
square-threaded screw should be, as frequently given in tech- 
nical books, \ $ of the pitch of the screw ; f #, as given in pre- 
vious tables, is more convenient, and also gives a little more 
wearing surface to the thread. The use of this table is so 
plain that it needs very little explanation. In the fourth column 
is the area of the outside diameter of the screw. In the fifth 
column, the sectional area of the screw at the bottom of the 
thread, which may be used in calculating the tensile and crush- 
ing strength of the screw. Subtracting the fifth column 
from the fourth gives the sixth column, which is the projected 
area of one thread ; this may be used in calculating the allow- 
able pressure on the thread, etc. The fourteenth column gives 
the tangential force which is required to act with a leverage of 
one foot in order to lift one pound by the screw if there was no 
friction. The fifteenth column gives the total tangential force 
required per pound of load when both load and friction are 
included. The sixteenth column gives the difference between 
the fourteenth and the fifteenth columns, and is the tangential 
force absorbed by friction alone. The coefficient of friction in 
both columns is assumed as 0.16. The last four columns in 
the table give the load or axial pressure which may be allowed 
on the screw corresponding to 200, 400, 600 and 1000 pounds 
pressure per square inch of projected area of screw thread 
when the length of the nut is twice the diameter of the screw. 



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MECHANICS. 



The table on page 313 was calculated by the following 
formulas : 

When friction is not considered : 
Force to balance load = Pitch in inches 



Pitch in inches 
12 X 2 77 7^4 

When both one pound of load and friction are considered, 

pitch in inches -f- middl e circum. X f \ ^ / middle radius \ 



Force = ( 



middle circum. — pitch in inches X f 



12 




calculations by table on preceding page. 

Example 1. 

A jack screw, as shown in 
Fig. 19, is iy 2 " diameter, three 
threads per inch. What tan- 
gential force is required to act 
with a leverage of 18 inches in 
order to lift 5000 pounds ? Co- 
efficient of friction in the thread 
is assumed as 0.16. Tangen- 
tial force absorbed in friction 
by the collar at a is assumed 
to be equal to force absorbed 
by friction in the thread of 
the screw, and may, therefore, 
be taken from the thirteenth 
column in the table. 

Solution : 

Tangential force per pound at 1 foot radius = 0.0133 
Tangential force absorbed by friction in collar — 0.0089 
Total force per pound of load at 1 foot radius = 0.0222 
The tangential force is acting with 18 inches leverage = 

\y z feet, and the load is 5000 pounds; therefore, the required 

force will be, 

F = °- 0222 X 5000 = 74 pounds. 

Example 2. 

A load of 16,000 pounds rests on a slide and is moved back 
and forth on a horizontal plane by a screw. The coefficient of 
friction between slide and plane is 0.1, and the screw should 
not be loaded with more than 400 pounds per square inch of 
projected area or thread. Find the suitable diameter of screw. 
If a pulley of 20-inch diameter is attached to the end of the 
screw, also find the tangential force required to act at the rim 
of the pulley in order to turn the screw. 



MECHANICS. 3 I 5 



Solution 



The coefficient of friction for the slide is 0.10, therefore the 
axial pressure on the screw will be 16,000 X Yio = 1600 pounds. 
The allowable force on a 1 X-inch screw will be found in the 
table to be 1742 pounds ; therefore, select a screw of 1% inches 
diameter and a length of nut of 2^ inches. Assuming the 
friction due to the reaction of the screw against its collar and 
bearing to be equal to the friction in the thread, and using the 
table, we have : 

Force per pound at one foot radius = 0.0112 

Force absorbed by friction in collar = 0.0074 

Total force per pound of load at one foot radius = 0.0186 

The leverage of a 20-inch pulley is 10 inches = 1 %2 foot, 
and the axial force is 1600 pounds ; therefore, the tangential 
force required at the rim of the pulley will be : 

F = 0-0186 X 1600 = 7 ds< 

10 /l2 

36.7 pounds is really the force required to keep the body in 
motion after it is started. To start the body from rest requires 
somewhat additional force, depending on the time used in over- 
coming its inertia. It is not certain that the friction due to the 
reaction of the screw against the collar is equal to the friction 
in the screw. It may be more or it may be less ; this will, to a 
certain extent, depend on the size of the collar, and also on the 
finish of its surfaces, its means of lubrication, etc. Therefore, 
instead of assuming this resistance to be equal to the friction of 
the thread as found in column 16, it may be calculated for 
each individual case by assuming a proper coefficient of friction 
and assuming that this friction acts as resistance at a radius 
equal to the middle radius of the collar. If a screw is acting 
under the circumstances illustrated in Fig. 18, there is no collar 
to absorb any of the force by friction ; but whenever the screw 
acts against a shoulder this friction must never be forgotten in 
calculation. Ball bearings may be used to very good advan- 
tage in the thrust collar on a screw. If a screw works a load 
continuously up and down, and the weight of the load always 
rests on the screw, it is necessary to be very careful and allow 
only a limited load on the screw (only a fraction of what is given 
in the table), because the pressure of the load always acts on 
the same side of the thread, and this is very disadvantageous 
for lubrication, as it does not give the oil a good chance to get 
onto the surfaces which rub against each other ; but when the 
screw works a slide with an alternate push and pull, the wear 
comes on both sides of the thread, which gives a good chance 
for lubrication, and an axial pressure of 400 pounds per square 
inch of projected area of bearing surface in the thread will be 



3 1 6 MECHANICS. 

safe, although, under certain circumstances, for instance, in a 
mechanism working continuously, such a load may be too much 
for the best results with regard to wear. 

For anything working like a jack-screw, when the diameter 
of the screw is over one inch, the load given in the last column 
is perfectly safe. It is impossible to give rules which will suit 
all cases ; the experience and judgment of the designer are the 
best guide with regard to the selection of the proper load. It 
may seem too much to use 0.16 as the coefficient of friction in 
the thread of the screw, but the author believes, from careful 
experiments made on common square-thread screws, as used in 
commercial machinery. — not made for experimental purposes, 
but for every-day use. — that this coefficient of friction is a safe 
average. It is well to remember that the surfaces of the thread 
on screws with cast-iron nuts do not always have the best of 
finish, and the nut especially is liable to be a little rough when 
new : therefore, this coefficient of friction may be a little greater 
than that found in screws in machinery when well lubricated 
and with surfaces smoothed down and glazed over from wear. 

The Parallelogram of Forces. 

A line may be drawn to such fig. 20. 

scale that its length represents a 
given force acting in the direction 
of the line. Another line is drawn 
to the same scale, from the same 
point of application, and its length 
represents another force acting in 
the same direction as this line. If 
these two lines are connected by 
two auxiliarv lines, a parallelogram 
is formed and the diagonal of the parallelogram will represent 
both the magnitude and the direction of the resulting force. 

Example. 

Let the lines a and b in Fig. 20 represent two forces acting 
in the direction of the arrows. Draw the lines to any scale, for 
instance. ^ inch to a pound : if the force represented by a is 64 
pounds, the line a will be 64 X -V = 4" long. If the force rep- 
resented bv b is 50 pounds, this line will be 50 X T V = &}&" 
long. Completing the parallelogram by drawing lines c and d, 
the diagonal, x, will indicate the magnitude and direction of the 
resulting force.' Suppose these two forces act in such direc- 
tions that when the parallelogram is completed and the diagonal 
drawn, it is. by measurement found to be 4%" long = ff; 
then the result of the two forces, a and b, is a force # of 76 
pounds. In many cases, the result of force and stress in ma- 
chinery and structures may very conveniently be obtained in this 
way with much less labor than by calculation, and with accuracy 
consistent with good, legitimate practice. 




HORSE-POWER. 317 

HORSE=POWER. 

The term horse-ftowe?; as applied in mechanical calcula" 
tions, is 33,000 foot-pounds of work performed per minute, or 
550 foot-pounds of work per second. 

To Calculate the Horse=Power of a Steam Engine. 

Rule. 

Multiply the area of piston in square inches by the mean 
effective steam pressure, and this by the piston speed in feet 
per minute, and divide this product by 33,000. The quotient is 
the horse-power of the engine. 

Formula : 

tt n 0.7854 Z) 2 X6X2sXn 
Horse-power — £. — — 

^ 33000 

D = Diameter of piston in inches. 

p =z Mean effective steam pressure in pounds per square 
inch. 

s = Length of stroke in feet. 

n = Number of revolutions per minute. 

Example. 

What is the horse-power of a steam engine of the following 
dimensions ? 

Cylinder, 20 inches diameter ; length of stroke, 3 feet ; 
number of revolutions per minute, 75 ; mean effective steam 
pressure in cylinder during the stroke, 60 pounds per square 
inch. 

Horse-power = 2 ° 2 X °- 78 54 >< 2.X 3 X 75 X 60 



Horse-power = 



33000 
314.16 X 450 X 60 



33000 
Horse-power = 257.04 

To Calculate the Horse=Power of a Compound or Triple 
Expansion Engine. 

Rule. 

Calculate the mean effective pressure of the steam (accord- 
ing to its number of expansions and initial pressure), and cal- 
culate the horse-power exactly as if it was a single cylinder 
engine of the same size as the size of the last cylinder. 

Another way is to take indicator diagrams of each cylinder, 
and calculate the power of each cylinder separately. 



318 HORSE-POWER. 

To Judge Approximately the Horse=Power which may be 
Developed by Any Common Single Cylinder Engine. 

Rule. 

Square the diameter of the piston in inches and divide 
by 2 : the quotient is the horse-power which the engine may 
develop. 

Note. — This rule gives the exact horse-power, if the prod- 
uct of the piston speed in feet and the average pressure per 
square inch in the cylinder is 21.000. 

Horse=Power of Waterfalls. 

Rule. 

Multiply the quantity of water in cubic feet falling in a 
minute by 62.5 : and multiply this by the height of the fall in 
feet: divide this product by 33.000. and the quotient is the 
horse-power of the waterfall. Or. multiply the quantity of 
water in cubic meters falling in a minute, by lOOO. and multiply 
this by the height of the fall in meters : divide the product by 
4500, and the quotient is the horse-power of the waterfall. 

Note. — The above rules give the gross power of the water- 
fall, but the useful effect of the fall is a great deal less and will 
depend on the construction of the motor. It may be only from 
40% to 50% of the natural power of the waterfall. 

Animal Power. 

Under favorable circumstances, a horse can perform 22.000 
foot-pounds of work per minute. For instance, a horse walking 
in a circle turning the lever in a so-called horse-power may 
exert a pull of 100 pounds, walking at a speed of 220 feet per 
minute. For the horse to work to advantage, the diameter of 
the circle ought to be at least 25 feet. 

Hauling a Load. 

The average speed when horses are used in hauling a load 
one way and returning without load the other way. allowing for 
necessary stoppages, may not be more than 175 feet per minute, 
and. in estimating, time must also be allowed for loading and 
unloading. Loads may vary from 1000 to 2000 pounds, accord- 
ing to the road. Commonly speaking, the force required to 
pull a loaded wagon on a good, level road increases in propor- 
tion to the load and decreases in proportion to the diameter of 
the wheels, and on soft roads it is less with wide tires than with 
narrow ones. The idea that a wagon having small wheels 
would be easier to pull up-hill than one having larger wheels is 
a fallacy. 



HORSE- POWER. 319 

Power of Man. 

A man may be able to do work at a rate of 4000 foot- 
pounds per minute ; for instance, in turning a crank on a crane 
or derrick, a force of 15 pounds may be exerted on a crank, 18 
inches long and, with 30 turns per minute, the work would be 
4228 foot-pounds per minute. 

Note. — In derricks, pulley blocks, jack-screws, etc., a large 
part of the expended power is consumed in overcoming friction. 

Power Required to Drive Various Kinds of Machinery. 

In the nature of the thing it is impossible from experiments 
on one machine to tell exactly what power it takes to run an- 
other similar machine, as there are so many different factors 
entering into the problem ; for instance, the speed and feed on 
the machine, the hardness of the stock it works on, the quality 
of the tools used, the kind of lubrication, etc. Therefore, such 
assertions are only approximations at the best. 

16-inch engine lathe, back geared, % horse-power. 

26-inch engine lathe, back geared, 1% horse-power. 

Planer, 22" x 22" x 6 feet, % horse-power. 

Planer, 32" x 32" x 10 feet, # horse-power. 

Shaping machine, 10-inch stroke, % horse-power. 

20-inch drill press, l / 2 horse-power. 

26-inch drill press, back gear, boring 

a 3-inch hole, using boring bar, 1 horse-power. 

Plain milling machines (Lincoln 

pattern, No. 2), 1% horse-power. 

Small Universal milling machines, ]4 horse-power. 

Circular saws (for wood), 24" di- 
ameter (light work), 0%. horse-power. 

Circular saws (for wood), 36" di- 
ameter (light work), 6 horse-power. 

Fan blower for cupola, melting four 

tons of iron per hour, 10 horse-power. 

Fan blower for five blacksmith fires, 1 horse-power. 

Drop hammer, 800 pounds, 8 horse-power. 

In machine shops and similar places, from 40% to 70% of 
the total power required is consumed in running the line shaft- 
ing and counter-shafts. An average of from 55% to 60% is 
probably the most common ratio. 

In exceptionally well-arranged establishments, under favor- 
able conditions, in light manufacturing it may be possible that 
only 30% of the power is consumed in driving line and counter 
shafting, and that 70% is used for actual work. 



::: :~n~ :? va:h:>ti?.y. 

SPEED OF MACHINERY. 

7 :.t z-tr-ZLz-L-rtlzzi— :: :irr2_2r 525 :iz;z: n:: :: ex:eef 
•A fee: per 12222: 2 :e ii'zeN: cT Aves ±ezuziber of revo- 
._r.:~s :er 22222:7 ::r zirzAir 52~~s :f fifferez: ziA~e:er5. 

TABLE No. 37. 



u :- 14 :: at 



N 



:•: ; ■ :•: . - .... ls»>: isoo aa ::_: 



Band Saws. 



Drilling Machines for Iron, 

7:; -lALAu: sreeL Ae s_rfi:e sreei :f 2 



Lathes. 



: \. fee: re: -inure 

>, lathes are usually 

_ feet on the slow- 

:er 25 Ae AA.e ■■ ... 

227 be fr:~ ! : '■' :: 
: :e rume-i is : 2: :f 



Planers. 

Ii5:-:::2 is rAnei 2: 2 sretf :: A :: ." fee: :e: ~A:::e 
; 2.: 12:2 A fee: s:eei 1 : :-r::t::22::r A rAner : 2272: 
:: re rim 2: ie2s: 222-ee rimes 25 :2s: :.. ; :: c:es :::~"iri. 

Hilling Machines 

Rotating cutters working on Bessemer steel or other mate- 
rials of about equal hardness usually h2 e 2 j^rface speee :: 



SPEED OF MACHINERY. 32 1 

about 40 feet per minute. Oil is used for lubrication. Cast- 
iron is milled without oil. 

Grindstones. 

When grindstones are used to grind steel and iron in manu- 
facturing, they work at a surface speed of 2000 to 2500 feet per 
minute, but grindstones for common shop use, to grind tools, 
chisels, etc., run at much slower speed. 

Emery Wheels and Emery Straps. 

Emery wheels and straps do good work at a speed of 5000 
to 6000 feet per minute, but all such high-speed machinery, 
especially grindstones and emery wheels, must be used very 
carefully and special attention paid to the strength, so that they 
will not break under the stress of centrifugal force. 

Calculating Size of Pulleys. 

TO FIND SIZE OF PULLEY ON MAIN SHAFT. 

Multiply the diameter of pulley on counter-shaft by its 
number of revolutions per minute, and divide this product by 
the number of revolutions of the main shaft, and the quotient is 
the diameter of the pulley on the main shaft. 

Example. 

A main shaft makes 150 revolutions per minute ; the counter- 
shaft has a pulley 9 inches in diameter and is to make 400 revolu- 
tions per minute. What size of pulley is required on the main 
shaft? 

Solution : 

Diameter of pulley — = 24 inches. 

150 

to find size of pulley on counter-shaft. 

Rule. 

Multiply the diameter of pulley on the main shaft by its 
number of revolutions per minute, and divide this product by 
the number of revolutions of the counter-shaft ; the quotient is 
the diameter of the pulley on the counter-shaft. 

Example. 

The pulley on a main shaft is 36 inches in diameter and it 
makes 150 revolutions per minute ; the counter-shaft is to make 
450 revolutions per minute. What size of pulley is required ? 

Solution : 

Diameter of pulley = 36 X 150 = 12 inches. 
' 450 



322 SPEED OF MACHINERY. 

TO FIND THE NUMBER OF REVOLUTIONS OF THE COUNTER 

SHAFT. 

Rule. 

Multiply the diameter of pulley on the main shaft by its 
number of revolutions per minute and divide this product by 
the diameter of pulley on the counter-shaft, and the quotient is 
the number of revolutions of the counter-shaft per minute. 

Example. 

The pulley on a main shaft is 24 inches in diameter and 
makes 150 revolutions per minute, and the pulley on the counter- 
shaft is 15 inches in diameter. How many revolutions per 
minute will the counter-shaft make ? 

24 X 150 

Number of revolutions = = 240 revolutions per minute. 

15 

To Calculate the Speed of Gearing. 

In calculating the speed of gearing, use the same rules as 
for belting, but take the number of teeth instead of the 
diameter. 

Example. 

The back gearing on a lathe consists of a gear and pinion 
of 8 pitch, 96 teeth and 82 teeth, and the other gear and pinion 
are 10 pitch, 120 teeth and 40 teeth. How many revolutions will 
the cone pulley make while the spindle makes one revolution? 

Solution : 

Cone pulley makes = — = 9 revolutions. 

* 32 X 40 

Efficiency of Machinery. 

Divide the energy given out by a machine by the energy 
put into the same machine ; multiply the quotient by 100, and 
the result is the per cent, of efficiency of the machine. 

Example. 

A dynamo requires 15 horse-power, but the electrical power 
given out is only 12 horse-power. What is the efficiency ? 
Solution : 

Efficiency = — X 100 = 80% 
15 

A steam engine is to develop 60 horse-power net. What 
will be the gross horse-power if the efficiency is 75% ? 
Solution : 

Gross power = 60 * 10 ° = 80 horse-power. 

(5 



CRANE HOOKS. 

CRANE HOOKS. 



3 2 3 




Fig. 2. 


"^ 










*^ 






Crane hooks, as shown in Figs. 1, 2 and 3, may be designed 
by the following formulas : 

P=D 2 D = VP~ 

P = Load in tons. 

D = Diameter of iron in inches. 
a=iy 2 B b = 3H£> c = \y*D 

e=±y 2 D f=2D g=l l A£> 

i=l%D j=%D t=%D 

1 = %D m = iy 2 D k — iy^D 

S = Standard screw of diameter r » = 

When a rectangular iron plate is substituted 
for a washer, the bearing surface of the plate 
against the wood should at least be equal to 
the area of the washer, calculated by the above 
formula. 

Chain Links. 

(See Figure 4.) 

D = Diameter of iron. 
L — ±y 2 to 5 D. 

B = sy 2 D. 

(For strength of chains, see page 222). 




324 



CRANES AND DERRICKS. 



CRANES. 



Cranes and derricks are 
machines used for raising and 
lowering heavy weights. In 
its simplest form, a crane con- 
sists of three principal mem- 
bers : The up-ight post, the 
horizontal jib and the diagonal 
brace. (See Fig. 5). The 
weight P will produce tensile 
stress in the jib, compressrve 
stress in the brace, and both 
compressive and transverse 
stress in the post. 



Fig. 5. 



Tension in jib = 



Compression in brace 



P X x 

y 

_ P Xz 

y 




Stress in the upper bearing 



_ PX h 



When the post is held at both ends, as in Fig. 5, it may, 
with regard to transverse strength, be considered as a beam of 
length i. fastened at one end and loaded at the other with a load 



equal to the force 



h X P 



The compression on the post caused by the load is equal 
to/ 3 . 

The downward pressure on the lower bearing is equal to 
the sum of the weight of the crane and the load which it 
supports. 



Proportions for a Two=Ton Derrick 



(Of the construction shown in Fig. 6) . 



Pulley blocks should be double-sheave (only single are 
shown in the cut). Circumference of manila rope, '?>% inches. 
Mast, 8X8 inches, 26 feet long. Boom, 7X7 inches, 20 feet long. 



CRANES AND DERRICKS. 



325 




Gear 



Large gear, 72 teeth, 1- 
inch circular pitch, 2-inch 
face. Small pinion, 12 
teeth, 1-inch circular pitch, 
2-inch face. Crank shaft. 
1)4 inches in diameter. 
Bearings, 2)4 inches long. 
Crank, 18 inches long, 
Drum, 7 inches in diame- 
ter, 24 inches long. Drum- 
shaft, 2% inches in di- 
ameter. The drum and 
large gear are fitted and 
keyed to the drum shaft 
and also bolted together, 
thereby relieving this shaft 
from twisting stress. 

The radius of the 
drum added to the radius 
of the rope makes four 
inches, and the force is 
multiplied five times by 
the double-sheave pulley 
block ; therefore, when 
the friction in thec ra nk 
mechanism is not consid- 
ered, the force required 
on the crank in order to 
lift 4400 pounds will be : 

F _ 4 X 12 X 4400 _ 33 nd nearly. 

18 X 72 X 5 

Thus, when two men are working the derrick (one at each 
crank), each man has to exert a force of 16)4 pounds, 
but, including friction, each man probably exerts a force of 20 
to 25 pounds, when the derrick is loaded to its full capacity. 

For very rapid work it is necessary to have four men (two 
on each winch-handle) to work the derrick, if it is kept loaded 
to its maximum capacity, but for ordinary stone work such a 
derrick is usually worked by two men. Stones as heavy 
as two tons are seldom handled, except where larger derricks 
and steam power are used. 

When the derrick is to be worked constantly, the limit of 
the average stress on the crank handle to be allowed for each 
man is 15 pounds. When working an 18-inch crank, 48 turns 
per minute, this corresponds to a force of 15 pounds acting 
through a space of a little over 220 feet = 3300 foot-pounds of 
work per minute = -^ horse-power. 

When the crank swings in a shorter radius a few more 
turns per minute may be expected, but experience indicates that 
an 18" radius is the most practical proportion. 



326 BELTS. 

BELTS. 

Oak-tanned leather is considered the best for belting. The 
so-called u short lap n is cut lengthwise from the middle of the 
back of the hide, where it has the most firmness and strength. 
Single belting more than three inches in width is about jV' 
thick, and weighs 15 to 16 ounces per square foot ; when less 
than three inches in width it is usually f^" thick and weighs 
about 13 ounces to the square foot. 

Light double belts, as used for dynamos and other ma- 
chinery having pulleys of comparatively small diameter, are 
abo-: - - :hick and weigh about 21 ounces per square foot. 
Double belting, as used for main belts, is a little heavier and 
weighs from 25 to 2S ounces per square foot. Belts as heavy as 
30 ounces per square foot are frequendy used, and are usually 
termed "heavy double." Large engine belts are sometimes 
made with three thicknesses of leather. 

Belts should be soft pliable and of even thickness. When 
a belt is of uneven thickness and has very long joints, so that 
it looks as if it was partly single and pardy double, it is very 
doubtful if it will do good service, for this is a sure sign that the 
thin and flimsy parts of the hide have been taken into the stock 
in making the belt. 

The ultimate tensile strength of leather belting is from 2600 
to -4-SOO pounds per square inch of section. Thus, a leather belt 
T V thick will break at a stress of 500 to 900 pounds per inch of 
width. 

The lacing of belts will reduce their strength from 50 to 60 
per cent: therefore, when practicable, belts ought to be made 
endless by cementing instead of lacing. 

A belt will transmit more power, wear better and last 
longer, if it is run with the grain side next to the pulley. 

Belts should never be tighter than is necessary in order to 
transmit the power without undue slipping : too tight belts cause 
hot bearings, excessive wear and tear, and loss of power in over- 
coming friction : but on the other hand, it is necessary to have a 
belt tight enough to prevent it from slipping on the pulley, be- 
cause if a belt slips there is not only a direct loss in velocity, but 
the belt will wear out in a short time : it is. therefore, very im- 
portant to use belts of such proportions that the power shall be 
transmitted with ease. 

Belts always run toward the side of the pulley which is 
largest in diameter (therefore pulleys are crowned, in order to 
keep the belt running straight). 

A belt will alwavs run toward the side where the centers of 
the shafts are nearest together. 

Open belts will cause two shafts to run in the same direction 



BELTS. 



327 



A crossed belt will cause the shafts to run in opposite direc- 
tions. If the distance between the shafts is short, crossed belts 
will not work well. A short belt will wear out faster than a 
long one. 

Very long and heavy belts should be supported by idlers as 
well under the slack as under the working side ; if not, the 
weight of a long belt will cause too much stress on itself and 
also cause too much pressure on the bearings, as well on the 
driver as on the driven shaft. Belts should never, when it can 
be avoided, be run vertically, as the weight of the belt always 
tends to keep it away from the lower pulley, thereby reducing 
its transmitting capacity; the longer the belt the worse this is. 
Belts are most effective when they are run in a horizontal direc- 
tion and, whenever possible, the lower part of the belt should be 
the working part, as the slackness in the upper part, by its 
weight, will cause the belt to lay around the pulley for a longer 
distance, and this will, in a measure, increase its transmitting 
capacity; but if the upper part is the working part, the slackness 
in the lower part tends to keep the belt away from the pulleys, 
and thereby reduces its transmitting capacity. 



Lacing Belts. 

Figure 1 shows a good way of lacing belts ; a is the side run- 
ning next to the pulley and b is the outside. Holes should be 
punched and not made by an awl, as punched holes are less lia- 
ble to tear. The lacing is commenced by putting each end of 
the lace through holes 1 and 2 from the side next to pulley, and 
then continuing toward the edges, both sides simultaneously, 



Fig, 



fV\ 



ft 




i 


f> 

s 


? 
it 







a 



making a double stitch at the edges and sewing back again un- 
til holes 1 and 2 are reached ; and, lastly, by drawing each end 
of the lace through x and y. Each stitch will be double, except- 



328 BELTS. 

ing the middle one. The holes x and y, where the ends of the 
lacing are finally drawn through for fastening, are made by the 
belt awl and should always be made small, and the lacing, if laid 
out rightly, always enters these holes from the inside of the belt ; 
after it is pulled through, a small cut is made in the lacing on 
the outside, which will prevent it from drawing back again, then 
the ends are cut off about %" long, as shown in the figure at.r 
and y. 1 1 is a bad practice to leave the lace-ends on the inside of 
belts, because they will then soon wear off, allowing the joint to 
rip. 

A 1-inch belt ought to have three lace-holes in each end. 
Length of lacing, 12 inches. 

A 2-inch belt ought to have three lace-holes in each end. 
Length of lacing, 18 inches. 

A 3-inch belt ought to have five lace-holes in each end. 
Length of lacing, 24 inches. 

A 4-inch belt ought to have five lace-holes in each end. 
Length of lacing, 32 inches. 

A 5-inch belt ought to have seven lace-holes in each end. 
Length of lacing, 40 inches. 

A 6-inch belt ought to have seven lace-holes in each end. 
Length of lacing, 48 inches. 

An 8-inch belt ought to have nine lace-holes in each end. 
Length of lacing, 60 inches. 

A 10-inch belt ought to have eleven lace-holes in each end. 
Length of lacing, 72 inches. 

A 12-inch belt ought to have thirteen lace-holes in each end. 
Length of lacing, 84 inches. 

Always have the row having the most holes nearest the end 
of the belt. 

Cementing Belts. 

When belts are cemented together, a 3-inch belt is lapped 
four inches and a 4-inch belt 4^ inches. In larger belts the lap 
is usually made equal to the width of the belt, but it may be 
made even shorter when the width of the belt is over 12 inches. 
The two ends are jointed together, so that the thickness is even 
with the rest of the belt. 

The American Machinist, in answer to Question No. 430, 
Dec. 5, 1895, says : " For leather belts take of common glue and 
American isinglass equal parts; place them in a glue pot and 
add water sufficient to just cover the whole. Let it soak 10 
hours, then bring the whole to a boiling heat, and add pure tan- 
nin until the whole appears like the white of an egg. Apply 
warm. Buff the grain of the leather where it is to be cemented; 
rub the joint surfaces solidly together, let it dry for a few hours, 
and the belt will be ready for use. For rubber belts take 16 
parts gutta percha, 4 parts India rubber, 2 parts common caulk- 
er's pitch, 1 part linseed oil ; melt together and use hot. This 
cement can also be used for leather." 



BELTS. 329 

Length of Belts. 

Small belts, such as 4 inches wide or less, will work well 
when the distance between the shafts is from 12 to 15 feet, larger 
belts when from 20 to 25 feet, and for large main belts 25 to 30 
feet distance is satisfactory. 

Horse=Power Transmitted by Belting. 

A single belt weighing about 15 ounces per square foot is 
capable of transmitting one horse-power per inch of width, 
when running at a speed of 800 feet per minute over pulleys of 
proper size, both of equal diameter. As one horse-power is 
33,000 foot-pounds of work per minute, this will make the tension 

, , , • • • 33000 

due to the power the belt is transmitting = o 00 =41)4 lbs. 

per inch of width, but the total tension in the belt is, of course, 
considerably more per inch of width, because the belt must be 
tight enough to prevent its slipping on the pulley. For belts 
lighter than 15 ounces per square foot it is better to allow 1000 
running feet per horse-power per inch of width of belt. For light 
double belts weighing 21 ounces per square foot, 600 running 
feet per horse-power per inch of width may be allowed. For 
double belts weighing 25 ounces per square foot, 500 running 
feet per horse-power per inch of width may be allowed. Hence 
the following formulas : 

For light single belts weighing less than 15 ounces per 
square foot, 

tt _ v X b HX 1000 

1000 ° v 

For single belts weighing 15 to 16 ounces per square foot, 
rr v X b H X 800 

800 ° ~ v 

For light double belts weighing about 21 ounces per square 
foot, 

v X b HX 600 

H - 600 *.- 7 v 

For double belts weighing about 25 ounces per square foot, 
v Xb i HX 500 



H ~ 50C 



H = Horse-power. 
b — Width of belt in inches. 

7/ == Velocity of belt in feet per minute, which will be di- 
ameter of pulley in inches multiplied by 3.1416 and by the num- 
ber of revolutions per minute, and the product divided by 12. 



S3° BELTS. 

Example 1. 

A double belt 10 inches wide, weighing 25 ounces per square 
foot, runs over 50-inch pulleys, making 240 revolutions per min- 
ute. How many horse-power will it properly transmit ? 

Solution: 

„ - . , ,_ , 50 X 3.1416 X 240 

Velocity of belt = ^ — 3141.6 ft. per minute. 

3141.6 X 10 
H = ~q — 62.8 horse-power. 

Example 2. 

One hundred horse-power is to be transmitted by a double 
belt weighing 25 ounces per square foot. The pulleys are 6(i 
inches in diameter and make 150 revolutions per minute. What 
is the necessary width of belt ? 

Solution : 

Pulleys of 66 inches diameter, running 150 revolutions per 
minute, will give a belt speed of 15 ° X 8 ' 1 ^ 16 X 66 = 2 591.8; 

sav, 2592 feet per minute. 

100 X 500 . , , , , , , , 

b — W\y> ~ 19.3 inches; thus, a double belt 20 

inches wide will do the work. 

Example 3. 

A light single belt 4 inches wide, weighing 13 ounces per 
square foot, runs over pulleys of 36 inches diameter, making 100 
revolutions per minute. How many horse-power may be trans- 
mitted ? 

Solution : 

36 X 3.1416 X 100 
V elocity of belt = t^ = 942.48 ft. per minute. 

The belt is a light single belt and its transmitting capacity 

4 X 942.48 
will be, H = 77w^ = 3.T6992, about §% horse-power. 

To Calculate Size of Belt for Given Horse=Power when 

Diameter of Pulley and Number of Revolutions 

of Shaft Are Known. 

The following formulas may be used for calculating belt 
transmission, and will give results approximately consistent 
with previously given rules, but they are more convenient for 
use, as the velocity of the belt does not need to be first calculat- 
ed, but the velocity of the belt must not exceed the practical 
limit. 



BELTS. 33I 

This formula will do for either single or double leather belts 
with cemented joints (no lacing), of any weight from 12 to 30 
ounces per square foot and of any width from one to thirty 
inches, when the pulleys are of suitable size to correspond with 
the thickness of the belt, and the diameter of both pulleys is 
equal or nearly so : 

d X n X b X w HX 50000 

■" — ^aaa?; d = 



50000 " — n X b X w 

H X 50000 H X 50000 H X 50000 

n ~ dX b X w b ~ dX nX w w ~ dXnXb 

H = Horse-power transmitted by the belt. 

d = Diameter of pulley in inches. 

n = Number of revolutions per minute. 

b = Width of belt in inches. 
<w = Weight of belt in ounces per square foot. 
50,000 is constant. 

Example. 

Calculate Example 2 by the above formula. 

Solution: 

100 X 50000 

b = (j(j x 150 X ">d ~ "^ mcnes ' wn i c h» f° r a ^ practical 
purposes, is the same as the result when calculated by the other 
rule. 

Wide and thin belts are unsatisfactory. It is far better 
when transmitting power to use double and narrow rather 
than single and wide belts. It is a very bad practice to run 
at too slow belt speed, and also to use pulleys of too small diam- 
eter. The smallest pulley for a light double belt should never 
be less than 12" in diameter, for a heavy double belt never less 
than 20" in diameter, and for a triple belt the pulley should not 
be less than 30" in diameter. 

To Calculate Width of Belt when Pulleys are of Unequal 

Diameter. 

W T hen the pulleys are of different diameters the belt will lay 
around the smallest pulley less than 180 degrees, and the trans- 
mitting capacity of the belt is correspondingly reduced. The 
pressure on the pulley due to the tension of the belt will vary as 
the sine of half the angle of contact, and the adhesion of the belt 
to the pulley will vary as the pressure : consequently, also, the 
transmitting capacity of the belt will vary as the sine of half of 
the angle of contact, but it is usually advisable in practice to 
allow a little more on the width of the belt than is called for 
by this rule. A practical rule is : 

First calculate the width of the belt by the above rules 
and formulas, as though both pulleys had the same diameter, 



33 2 BELTS. 

then multiply the result by the following constants, according to 
the arc of contact between the belt and the small pulley. 

When the arc of contact between the belt and the small 
pulley is 90° multiply by 1.60. 

100° " " 1.45 140° multiply by 1.15 

110° " " 1.35 150° " " 1.10 

120° " " 1.25 160° " " 1.06 

130° " " 1.20 170° " " 1.04 

Example. 

The pulley on a dynamo is 15" in diameter, and it makes 
1200 revolutions per minute. The driving pulley is so large that 
the belt only lays around the dynamo pulley for a distance of 
150 degrees. What is the necessary width of a light double belt, 
weighing 21 ounces per square foot, when it takes 40 horse-power 
to run the dynamo ? 

Solution : 

If the arc of contact had been 180 degrees the belt would 
40 X 50000 
be b ■=■ 1900 X 15 X -21 = ^ mc hes wide, but as the arc 

of contact is not 180 degrees, but only 150 degrees, this width 
is multiplied by the constant 1.10, as given in the preceding 
table. Thus, the width of the belt will be 5.3 X 1.1 — 5.83 
inches or, practically, a belt six inches wide is required. 

When belts are running in a horizontal direction, 
and the driven pulley and the driver are of equal diameter 
and finish, the belt will always, when overloaded, commence to 
slip on the driver, and when pulleys are of unequal size it is 
always more favorable for the belt when the driving pulley is 
the larger than when vice versa. 

To Find the Arc of Contact of Belts. 

Make a scale drawing of the pulleys and the belt, and 
measure the arc of contact from the drawing by means of a 
protractor, or the arc of contact in degrees on the small 
pulley for an open belt may be calculated by the formula : 

Cosine of half the angle = — ■ 

R = Radius of large pulley in inches. 
r ■=■ Radius of small pulley in inches. 
/ = Distance in inches between centers of the shafts. 

Example. 

The distance between centers of two shafts is 16 feet ; the 
large pulley is 60 inches and the small pulley is 20 inches 
in diameter. What is the arc of contact of the belt ? 



BELTS. 333 

Solution : 

16 feet = 192 inches. 

60 inches diameter === 30 inches radius. 

20 inches diameter = 10 inches radius. 

Cos. of half the angle = ( 30 ~ 10 ) — 0.104 
& 192 

In tables of natural cosine (page 158), the corresponding 
angle is found to be 84 degrees, very nearly ; thus, the angle 
for arc of contact will be 2 X 84 = 168 degrees on the small 
pulley. On the large pulley the arc of contact will be 360 — 
168 = 192 degrees. 

For a crossed belt the arc of contact is always the same on 
both pulleys, and it may be calculated by the formula : 

Cos. of half the angle = — — -, — 

R = Radius of large pulley. 
r — Radius of small pulley. 
/ = Distance between centers. 

Example. 

What will be the arc of contact for the belt on the pulleys 
in the previous examples if belt is run crossed instead of open ? 

Solution : 

Cosine of half the angle = — _° ~*~ 10 = — 0.208 ; the 

192 

corresponding angle will be 180 — T7 = 103 degrees, and the 
arc of contact will be 103 X2 = 206 degrees. 

Pressure on the Bearings Caused by the Belt. 

Approximately, the pressure on the bearings caused by the 
belt may be considered to be three times the force which the 
belt is transmitting. Therefore, the pressure may be calculated 
by the formula : 

p _ 3 X 33000 X H 

v 

P = Pressure on the bearings due to pull of belt. 

H =. Number of horse-power transmitted by the belt. 

v — Velocity of belt in feet per minute. 

Example 1. 

A belt is transmitting 60 horse-power and its velocity is 900 
feet per minute. What is the pressure in the bearings due to 
the belt ? 



334 belt; 

Solution : 

p _ 3 X 33000 X 60 _ nfk , 

F — — = 6600 pounds. 

yuu 

Example 2. 

Suppose the diameters of the pulleys are increased until a 
belt speed of 3000 feet per minute is obtained. What will then 
be the pressure in the bearings caused by the belt when trans- 
mitting 60 horse-power ? 

Solution : 

p _ 3 X 33000 X 60 . OOA , 

r* = = 1980 pounds. 

3000 ^ 

By the above examples it is conclusively shown what a 
great advantage there is in using pulleys so large in di- 
ameter that proper belt speed is obtained. (See velocity of 
belts, page 337). 

The approximate pressure may also be very conveniently 
obtained from the width of the belt, thus: For light single 
belts, allow 1000 feet of belt speed per horse-power transmitted 
per inch of width of belt. The effective pull in such a belt will 
be 33 pounds per inch of width, and the pressure on the bearings 
due to the belt will accordingly be 33 X 3 = 99 pounds per inch 
of width of belt. For convenience, say 100 pounds pressure in 
the bearings per inch of width of such belts. For belts where 
800 running feet are allowed per horse-power per inch of width 
of belt, this reasoning will give a pressure on the bearing equal 
to 123^ pounds per inch of belt. For convenience, say 125 
pounds pressure in the bearings per inch of width of such 
belts. For belts where 600 running feet are allowed per horse- 
power per inch of width, the pressure in the bearing is equal 
to 165 pounds per inch of width of belt, and where the belt is 
so heavy that only 500 feet of belt speed per horse-power per 
inch of width is allowed, the pressure in the bearings will be 
198 pounds per inch of width. A good, practical rule, which 
can very easily be remembered, is, (when belts are in good order 
and have the proper size and the proper tension) : 

Multiply weight of belt in ounces per square foot by eight 
times the width of the belt in inches, and the product is 
approximately the pressure in pounds upon the bearings caused 
by the belt. 

Example. 

A belt is calculated with regard to the horse-power it has 
to transmit under a given velocity, and found to be 8-inch 
double belting, weighing 25 ounces per square foot. What pres- 
sure will it cause on the bearings when working at proper 
tension ? 



BELTS. 335 

Solution, by the last rule : 

? = 8 X 25 X 8 = 1600 pounds. 

Solution, by the first rule: 

At a speed of 3000 feet per minute such a belt will transmit 

3000 X 8 

— 500 — = 48 horse-power, and calculating the pressure by the 
formula : 

P = 



3 X 33000 X H 



p 3 X 33000 X 48 = 1584 ^ 

3000 

Both rules give nearly the same result, and one is just as 
correct as the other, as all such figuring is nothing more than 
approximation at the best. The pressure on the bearings may 
be a great deal more than calculated above. Sometimes the 
pulleys are roughly made, belts are poor, and consequently the 
coefficient of friction between belt and pulley is small, and as 
the belt has to be a great deal tighter in order to do the work, 
the pressure on the bearing will be greatly increased. Very 
frequently, from pure ignorance or carelessness, belts s are made 
very much tighter than necessary, and enormous sums of money 
may be wasted in this way in large factories, as the steam 
engines, at the expense of the coal pile, have to furnish power 
not only to do the useful work, but also to overcome all the 
friction produced by such over-strained belts, hot bearings, etc. 
A belt will transmit more power over a good, smooth pulley 
than over a rough one. When pulleys are covered with leather 
a belt will transmit about 25% more power than it will when 
running over bare iron pulleys, and in transmitting the same 
power a much slacker belt may be used, thereby reducing the 
friction in the bearings. 



Special Arrangement of Belts. 

By the use of suitable guide pulleys it is possible to connect 
with belts shafts at almost any angle to each other. But 
experience is required and care must be exercised to do it suc- 
cessfully. When guide pulleys are used in order to change the 
direction of a belt, always remember that when the belt is run- 
ning the most pressure is thrown on the pulley guiding the 
working part of the belt. This pulley is, therefore, very liable 
to heat in its bearings, if not designed to have bearing surface 
enough and also to have proper means for oiling. 



3tf 



BELTS. 



Fig. 2 shows an arrangement 
by which the direction of motion 
of two shafts may be reversed, 
when the distance between the 
shafts is too short for the use of a 
crossed belt or when a crossed 
belt, for any other reason, cannot 
be used. 

Suppose pulley A to be the 
driver and to run in the direction of the arrow. C and D are 
guide pulleys, and the motion of the driven shaft B is in the op- 
posite direction to the shafts. In this case the guide pulley C is 
on the working part of the belt, and is the one to which special 
attention must be paid in regard to heating. If the direction 
of shaft A is reversed, guide pulley D will be on the working 
part of the belt. 




Crossed Belts. 

If the distance between A and B (Fig. 2) had been long 
enough, it would have been preferable to reverse the motion of 
B by means of a crossed belt, instead of by the arrangement 
shown in Fig. 2. 

Crossed belts do not work well when running on pulleys 
small in diameter as compared to the width of the belt. 

Too short distance between the shafts must be avoided. 

Wide crossed belts are very unsatisfactory; therefore, 
instead of running one wide crossed belt it is preferable to use 
two belts, each of half the width, and run them on two separate 
pairs of pulleys. Such belts should be of equal thickness, and 
the pulleys should be crowned, well finished and of correct 
size, so that each belt will do its share of the work. 



Quarter=Turn Belts. 

Fig. 8 shows a so-called quarter- 
turn belt, used to connect two shafts 
when running at an angle and laying 
in different planes. The principal 
point to look out for is to place the 
pulleys (as shown in Fig. 3) so that 
the belt runs straight from the de- 
livering to the receiving side of each 
pulley. 

The pulleys shown in Fig. 3 are 
set right for belts running in the 
direction of the arrows. If the mo- 
tion is reversed, the belt will run off 
the pulleys. 



FIG. 3. 





BELTS. 337 

Angle Belts. 

The belt arrangement shown in Fig. 4 is usually called an 
angle belt, and is used to connect two shafts at an angle. Either 
one, A or B, may be the driver, and 
there are two guide pulleys (one for 
each part of the belt at C), one of which, 
of course, is on the driving part of the 
belt. 

Crossed belts, quarter-turn belts, 
and angle belts must never be wide and 
thin ; much better results are obtained 
by narrow, double belts than by wide, 
single ones. 

Angle belts and quarter-turn belts 
are frequently bothersome contrivances. 
Their running is sometimes improved 
by making a twist in the belt when 
joining its ends ; that is, lacing the flesh side of one end and the 
hair side of the other end on the outside. This will prevent one 
side of the belt from stretching more than the other. 

Slipping of Belts. 

Owing to the elasticity of belts, there must always be more 
or less slip or " creep " of the belts on the pulleys. Under 
favorable conditions it may be as low as 2%, but frequently the 
slip is more. Therefore, if two shafts are connected by belts, 
and both should have very nearly the same speed, the diameter 
of the driver should be at least 2% larger than the diameter of 
the driven pulley. When the driver is comparatively large in 
diameter and the driven pulley is small, it is advisable to have 
the driver from 2 to 5% over size, in order to get the required 
speed. 

Tighteners on Belts. 

If tighteners are used they should always be placed on the 
slack part of the belt. 

Velocity of Belts. 

Belts are run at almost all velocities from less than 500 to 
5000 feet per minute, but good practice indicates that whenever 
possible main belts having to transmit quantities of power are 
run most economically at a speed of 3000 to 4000 feet per min- 
ute. At a higher speed both practice and theory seem to agree 
that the loss due to the action of the centrifugal force in the belt 
when passing around the pulley, and that the wear and tear is so 
great when the speed is much over 4000 feet per minute that 
there is not much practical gain in increasing the speed. But, 
as a general rule, whenever possible the higher the belt speed 
the more economical is the transmission as long as the belt 
speed does not exceed the neighborhood of 4000 feet per minute. 



$$$> BELTS. 

Oiling of Belts. 

Belts should be kept soft and pliable and are. therefore, 
usually oiled with either neat"s-foot oil or castor oil. Too much 
oiling is hurtful, but the right amount of oiling at proper rimes 
is very beneficial to the action of the belt and will prolong its 
utility to a great extent. 

Remarks. — All previous rules for calculating belting are 
founded upon good, legitimate practice, but are only offered as 
a guide, as no rule can be given which will fit all cases. 

For instance, a belt may be amply large to transmit 
a given horse-power when running in a horizontal direction, 
but it may fail to do the same work if running in a vertical di- 
rection. A belt may be large enough to do its work when run- 
ning in a vertical direction over pulleys of unequal size with 
the large pulley on the lower shaft, but it may fail to do the 
same work satisfactorily with the large pulley on the upper shaft 
and the small pulley on the lower one. 

Leather belts should not be used where it is damp or wet 
but rubber belting will usually give good service in such 
places. 

For information regarding rubber belts, see manufacturers' 
catalogues. 



WIRE ROPE TRANSMISSION. 

Transmitting power by wire ropes running at a high speed 
over grooved pulleys, or u telodynamic transmission." as it is 
also called, is the invention of the brothers Hirn of Switzerland. 
For long distances this mode of transmission is far cheaper than 
leather belting or lines of shafting. Fig. 1 shows a section of a 
pulley as used for this kind of transmission : a is an elastic fill- 
ing, usually made from leather cut out and packed in edgewise. 
The groove is made wide, so that the rope will rest entirely 
against the packing and not touch the iron. This is different 
from transmission with hemp rope, which is made to wedge into 
the groove of the pulley. 

The diameter of the pulley in the groove, where the wire 
rope runs, ought to be at least 150 times the diameter of the 
rope : the larger the better, so long as the velocity of the rope 
does not exceed 5000 feet per minute. The pulleys must run 
true and be in balance and in exact line with each other, and the 



WIRE ROPE TRANSMISSION. 



339 



shafts must be parallel. The distance between shafts should 
never be less than 60 feet and should preferably be from 150 to 
400 feet* For distances longer than 400 feet, either carrying 
pulleys or intermediate jack shafts are generally used, although 
spans as long as 000 feet or more have been used, but only when 
it is possible to give the rope the proper deflection without its 
touching the ground. Usually the speed is from 3000 to 6000 
feet per minute. Higher speed would be dangerous from the 
stress in cast-iron wheels due to centrifugal force. 









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Tightening pulleys should not be used, because if the distance 
between centers of shaft is too short to give the proper tightness 
to the rope without a tightening pulley, wire rope transmission 
is not the form best adapted to the circumstances. Guide 
pulleys or idlers should be avoided as much as possible, but 
when necessary they should be as carefully made and put up as 
the main pulleys, and they ought not to be less than half the 
diameter of the main pulley if on the slack part, but of the same 
size if they are on the tight part of the rope. Wire rope for 
transmission is usually made from the best quality of iron, 
has seven wires to a strand and consists of six strands laid 
around a hemp core in the center of the rope. The diameter 
of the wire rope is from nine to ten times the diameter of each 
single wire. 

Never use galvanized rope for power transmission, but pre- 
serve the rope by painting with heavy coats of linseed oil and 
lampblack. 



* When distance between shafts is less than 60 feet, leather belts are prefer- 
able to wire rope. 



340 WIRE ROPE TRANSMISSION. 

Transmission Capacity of Wire Ropes. 

A one-inch rope running 5000 feet per minute is capable of 
transmitting 200 horse-power. The transmitting capacity of the 
rope is in proportion to the square of its diameter, and the power 
transmitted by the rope when the velocity- is less than 5000 feet 
per minute is practically in proportion to its velocity* Hence 
the formula : 

H= ^ XfX 2Q0 which reduces to H - 0.04 X d* X V 
5000 

H= Horse-power transmitted. 
d= Diameter of rope in inches. 
V= Velocity of rope in feet per minute. 

Example. 

How many horse-power may be transmitted by a wire rope 
% inch in diameter running over proper pulleys at a velocity of 
2500 feet per minute ? 

Solution : 

H— 0.04 X y z X % X 2500 = 25 horse-power. 
The pressure on the bearings will not be less than three 
times the force transmitted, and may be calculated thus : 

Pressure on bearings = 3 X horse-power X 33000 

\ elocity in feet per min. 

Example. 

What will be the least pressure in bearings for a wire rope 
transmitting 150 horse-power at a velocity of 5000 feet per 
minute ? 

Pressure on bearings = 3 X 15 ° X 330 ° = 2970 pounds. 

5000 P 

If there is one bearing on each side at an equal distance 
from the pulley, the pressure on each bearing will be 2 V ° = 
1485 pounds. This is the calculated pressure, and represents 
what the pressure should be. but it is not certain that this is the 
actual pressure. It may be greatly increased by having the rope 
too tight, t 

* When the velocity of the rope exceeds 6000 feet per minute the stress caused 
by centrifugal force when the rope is bending around the pulley considerably 
reduces its transmitting capacity. This loss increases very fast above this speed, 
because the centrifugal force increases as the square of the velocity. It is veiy 
doubtful if there is practically any gain to run wire ropes at a speed exceeding 6000 
feet per minute when wear and tear, loss due to centrifugal force, etc., are 
considered. 

t Sometimes a pulley is put on the free end of a line of shafting projecting 
through the wall and drawn by a wire rope outside the shop; this will do only 
when a comparatively small amount of power is to be transmitted. 



WIRE ROPE TRANSMISSION. 



341 



The tension of the rope may be calculated from its de- 
flection when at rest (see Fig. 2), and for a rope running hor- 
izontally the usual formula is : 




P 



_WX Va 2 -f & 



P = 



a 



(very nearly) 



P = Force in pounds at f. 

W = Weight of rope in pounds from d to f, which is half 
the span. 

b = Half the span in feet. 

a = Twice the deflection in feet. 

Note. — (See Fig. 2.) If the length of the line a represents 
the weight of the part of the rope from dtof. the length of the 
line x represents the tension in the rope at/"; therefore the ten- 
sion will be as many times the weight as the length of line a is 
contained in the line x. 

Example. 

The horizontal distance between two pulleys is 200 feet ; 
when standing still the deflection in a wire rope of 7 /&" 
diameter is 5 feet. What is the tension in the rope ? 

Solution : 

In Table No. 38 the weight of J4" wire rope is given as 1.12 
pounds per foot; therefore, 100 feet of ys" rope will weigh 112 
pounds. 



P — 



112 X \/l00 2 + 10- _ 112 X 100.5 



10 



10 



= 1125. G pounds. 



This is the tension in each part of the rope ; therefore the 
force against the pulley, due to the weight of the rope, is 1125.(5 
X2 = 2251.2 pounds. If this is supported by a bearing on each 
side of the pulley, the pressure on each bearing, if both are the 
same distance from the pulley, will be 1125. G pounds. 



34- WIRE ROPE TRANSMISSION. 

The tension is increased by reducing the deflection. For 
instance, if the deflection is reduced to -A feet the tension on the 
rope will be. 

P = 112 X V'lOtP-hS 2 

P = n - X^ 100.3 _ 1404 2 polmds 

Thus, the tension might be increased to any amount within 
the ultimate breaking strength of the rope. 

Deflection in Wire Ropes. 

When the rope is in motion the deflection will increase on 
the slack side and decrease on the tight side : therefore, if the 
span is long the rope may touch the ground when running if the 
pulleys are not placed on sufficiently high towers. There is 
really nothing else which, within practical limits, determines the 
length of the span, which may just as well be 1000 feet, or even 
more, providing the proper deflection can be given to the rope 
without touching the ground. When possible the lower part of 
the rope should be the working side, but in a long span this is 
impossible, because, when running, the lower part of the rope 
would be tight and the upper part slack, causing the two parts 
of the rope to strike together, which must never be allowed. 
When the length of the span exceeds 35 times the diameter of 
the pulleys it is safest to have the upper part of the rope the 
working side and the lower part the slack side. 

When the lower part of the rope is the slack side, the least 
space allowable for the slack of the rope at the center of the 
span will (when the rope is as tight as given in Table No. 38), be 
obtained by the formula : 

Distance = 0.00015 X (span) 2 

but. to allow for contingencies, it is better to have more room. 
When the lower side of the rope is the tight side, the rope will 
be clear from the ground when running if the space is 0.0001 X 
(span -. The deflection in the rope when standing still which 
will produce a pressure on the bearings and give tension enough 
to transmit the horse-power given in Table Xo. 38, may be cal- 
culated approximately by the formula : 

d = 0.00009 x r 2 

d = Deflection in feet. 

/ = Distance between pulleys in feet. (See Fig. 2). 

Example. 

The distance between the pulleys being 400 feet find the 
greatest allowable deflection in the rope, when standing still, in 
order to transmit the horse-power given in Table Xo. 38. 



WIRE ROPE TRANSMISSION. 



343 



Solution : 

d = 0.00009 X 400 X 400 = 14.4 feet. 

When the rope is new it is always put on with more 
tension than is necessary to transmit the power, because new 
rope will stretch. It is, therefore, very important when de- 
signing such transmission to calculate the maximum pressure 
which the rope will exert on the bearings when put on with the 
least deflection ever wanted, and calculate size of bearings and 
shafting for pulleys according to this stress, with due considera- 
tion not only for strength but also for heat and wear. (See page 
360 and page 367.) The correct amount to allow for stretch will 
vary with different kinds of rope and also with the tempera- 
ture. If a rope is spliced on a warm summer day it must be 
made slacker than if it was spliced on a cold winter day, as the 
length of the rope will be changed considerably by the difference 
in temperature ; the only guide is practical experience and good 
judgment. As a general rule, it may be safe to allow about half 
of the deflection as previously calculated when splicing a new 
rope, provided that the shafts and bearings are constructed so 
as to allow such tension. The rope is always strong enough. 
The splicing of the rope should be done by a man ex- 
perienced in that kind of work. The splice itself is usually 
made at least 240 times the diameter of the rope. 

TABLE No. 38.— Giving SuitabIe=Sized Pulleys for Different Sizes of 
Wire Rope, Weight of Rope, Horse=Power which Different Sizes 
of Wire Rope flay Transmit at Different Velocities, the least Stress 
at which it may be done and the Least Corresponding Pressure on 
the Bearings ; also, the Ultimate Average Strength of Wire Rope. 



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344 wire r ope transmission. 

Example. 

From a shaft running 150 revolutions per minute 100 horse- 
power is to be taken off by a wire rope. The velocity of the rope 
is to be 5000 feet per minute. What size of pulley and rope 
will be required ? 

Solution : 

In Table No. 38 it will be found that a ^-inch wire rope, run- 
ning at 5000 feet per minute, is capable of transmitting 112 horse- 
power ; thus, select a ^-inch rope. The diameter of the pulley 

5000 
will be i5Q x 3 1416 = ^'^ ^ eet * * n ^ e ta ^ e ** w ^ ^ e 
found that a 10-foot pulley is the smallest advisable to run 
with a ^-inch wire rope, therefore the pulley 10.6 feet in diame- 
ter is within the requirements. The next step is to calculate the 
pressure on the bearings. In the table it is found that the least 
pressure due to the transmission of 112 horse-power is 1908 
pounds. This cannot be used in calculating sizes of shafts and 
bearings, but use the maximum pressure, which is calculated ac- 
cording to the allowable deflection in the rope, as explained 
on page 341. Also consider weight of pulley and shaft, then 
calculate size of shaft and bearings, with due consideration to 
strength, stiffness, wear, heat, etc. ( See pages 360-367.) 

Transmission of Power by Manila Ropes. 

Manila ropes are used more or less for transmission of 
power. In this country one continuous rope, going back and 
forth in separate grooves over the pulleys several times, is fre- 
quently used, and a tightening arrangement is placed on one of 
the slack parts, which automatically keeps the rope at the 
proper tension, regardless of changes due to weather or stretch 
due to wear. This arrangement has its advantages in keeping 
the rope at more even tension than is possible with the Euro- 
pean system, but the disadvantage is that if a break occurs the 
transmission is entirely disabled until it is repaired. The Euro- 
pean practice is to use several single ropes running in separate 
grooves side by side on the same pulley. This has the advan- 
tage that if one of the ropes should break it is usually possible 
to run undisturbed until there is a chance to repair it, because 
it is always advisable to have margin enough in the transmission 
capacity of the ropes so that the shaft will run satisfactorily, 
even if one rope is taken off. The disadvantage of this system 
is the difficulty in keeping all the ropes at equal tightness and 
getting them to pull evenly. 

Fig. 3 shows the usual shape of pulley used for manila 
ropes, which may be made from either wood or iron. The 
European practice is to use iron, but whichever material is used 
it is very important to have the sides of the grooves care- 
fully polished, as the rope rubs on the sides in entering and 



MANILA ROPE TRANSMISSION. 



345 



leaving the pulley and will wear out in a short time if the 
pulley is left as it comes from the lathe tool. Sand and blow- 
holes must also be avoided. The angle of groove is usually 45°, 
and the rope is made to wedge into it, as shown in Fig. 3. 

The usual shape of grooves for guide pulleys is shown in 
Fig:. 4. 



Fig. 3. 



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FIG. 4. 




The best speed for ropes is from 1500 to 5000 feet per 
minute. ' When the velocity of the rope exceeds 6000 feet per 
minute the loss, due to the centrifugal force, is so great that it 
will hardly pay to increase the velocity. The diameter of the 
pulleys ought to be at least 50 times the diameter of the rope. 



Transmission Capacity of Manila Rope. 

A manila rope two inches in diameter, running over 
properly-shaped pulleys at a speed of 5000 feet per minute, is 
capable of transmitting 50 horse-power. The transmitting 
capacity of the rope is in proportion to the square of its 
diameter, and the power transmitted by the rope is in propor- 
tion to its velocity 7 ; therefore, a one-inch rope, running 5000 feet 
per minute, will transmit 12^ horse-power, and the formula 
will be : 

Horse-power = d% X " X 12 ' 5 
- 5000 

which reduces to 

Horse-power = 0.0025 X d 2 X v 
d = Diameter of rope in inches. 
v — Velocity of rope in feet per minute. 



346 



MANILA ROPE TRANSMISSION. 



Example. 

What horse-power may be transmitted by a manila rope 
\y 2 inches in diameter, running over nine-foot pulleys at a speed 
of 150 revolutions per minute ? 

Solution : 

Nine-foot pulleys, running 150 revolutions per minute, give 
the rope a velocity of 3.1416 X 9 X 150 — 4241 feet per minute, 
and the horse-power transmitted will be : 

H-P — 0.0025 XlKXl^X 4241 

H-P — 0.0025 X 2% X 4241 

H-P = 23.85 : practically, 24 horse-power. 

Weight of Manila Rope. 

The weight of one foot of manila rope of one-inch diameter is 
T % pound ; therefore, the weight per foot of any size may be 
calculated approximately by the formula : 

IV =d 2 X 0.3 
</= Diameter of rope in inches. 

W = Weight of rope in pounds per foot. 

Example. 

What is the weight of 360 feet of manila rope of l>^-inch 
diameter ? 

Solution : 

Weight of 360 feet = 0.3 X 360 X1^X1^= 243 pounds. 

TABLE No. 39, 

Giving the Weight of Rope in Pounds per Foot, Driving Force in 
Pounds, and Corresponding Horse= Power Transmitted at Differ= 
ent Velocities. 



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w 3 


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3 


!> 


ja 


3 


fa S fa S 


fa C 


fa 7 


fa 


fa - 


fa = 


fa ~ 


Sft 


6 


bp 


]> 


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© i) 


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o 3 


O 1) 


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fr 


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CO 


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IO 


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2 


X 


0.0T5 


21 


0.94 


1.25 


1.56 


1.87 


2.18 


2.50 


3.12 


3.75 


2X 


H 


0.12 


33 


1.45 


1.94 


2.24 


2.90 


3.39 


3.87 


4.48 


5.81 


2^ 


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47 


2.11 


2.81 


3.52 


4.82 


4.92 5.62 


7.03 


8.44 


s/ 2 


l 


0.30 


83 i 


3.75 


5 


6.25 


7.50 


8.75 10 


12.50 


15 


4% 


IX 


0.47 


132 


5.86 7.81 9.77 


11.72 13.73 15.02 


19.23 


23.44 


5 


i# 


0.67 


186 


8.4411.25 


14.06 


16.87 19.69 22.50 


28.12 


33.75 


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0.92 


255 


11.48 


15.31 


18.31 


22.97 26.79 30.62 


36.61 


45.94 


8 


2 


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330, 


15 


20 


25 


30 35 ,40 


50 


60 



MANILA ROPE TRANSMISSION. 347 

The transmitted horse-power, as given in Table No. 39, is 
calculated by the formula, 

H-P — 0.00025 X d 2 Xv 

Example. (Showing application of Table No. 39.) 

What size of rope is required to transmit 50 horse-power 
when three independent ropes are used, running over the same 
pulley at a velocity of 4000 feet per minute ? 

Solution : 

It is always advisable to select ropes having sufficient trans- 
mitting capacity to continue the transmission undisturbed, even 
if one rope breaks ; therefore, select ropes of such size that two 
ropes will transmit nearly 25 horse-power each. In Table No. 
39 it is found that a manila rope 1% inches in diameter, running 
4000 feet per minute, will transmit 22.5 horse-power. Thus, this 
will be the size of rope to use. The small pulley in the trans- 
mission must not be less than five feet in diameter. (See Table 
No. 39.) 

The pressure on the bearings, due to tension of the rope, 
will not exceed three times the driving force, because manila 
ropes run comparatively slack, as the adhesion to the pulley 
does not depend so much on the tightness of the rope as it does 
on its wedging into the groove in the pulley. The driving 
force of manila rope of lX-inch diameter is given in the table 
as 132 pounds; therefore, the pressure due to one rope will be 
3 X 132 := 396 pounds, and the pressure due to three ropes will 
be 3 X 396 = 1188 pounds ; besides this, the weight of the shaft 
and the pulley should be considered when calculating the size of 
shaft and bearings, with due consideration for strength, stiffness, 
wear, heat, etc. (See page 367.) 



Preservation of Manila Rope. 

The life of the rope is prolonged by slushing once in a 
while with tallow mixed with plumbago. The rope will not only 
wear on the outside but also within itself, because the fibers 
chafe on each other as the rope bends over the pulleys ; hence 
the preference for pulleys of large diameter. If the rope is not 
specially prepared for transmission purposes, it ought to be 
soaked in a mixture of plumbago and melted tallow when new, 
before it is used. There is on the market manila rope especial- 
ly manufactured for transmission purposes, having the fibers 
treated with plumbago and tallow, and, whenever obtainable, 
such rope should be used, as it will last much longer and give 
much better service than ordinary manila rope. 



348 



PULLEYS. 



PULLEYS. 

The following empirical rule gives arms of nice shape and 
good proportions : 

When the diameter of the pulley is at least 4 times its face, 
use 6 arms for pulleys from 12 to 60 inches. For a 12-inch pulley 
make the arms 1 % inches wide at the hub and add T ] g of an inch 
to the width of the arm for each inch the pulley is increased in 
diameter. 



Formula : 



h — 



D — Yl 
16 



+ 1X 



h = Width of arm in inches projected to center of hub. 

(See Fig. 1.) 
D = Diameter of pulley in inches. 

Example. 

Find width of arms at the hub for a 60-inch pulley. 

Solution : 

60 — 12 , 
h — — Tg -h IX = 4X inches. 

The width of the arm at the rim should be three-fourths of 
the width at the hub, and the thickness should be one-half of the 
width for arms with segmental sections (seejj/ Fig. 1) and four- 
tenths of the width for elliptical form of section ; (see x Fig. 1.) 
For double belts multiply these diameters by 1.3. 

fig. 1. Large, well-round- 

WV ed fillets must be used 

where the rim and arms 
meetat;/z. (SeeFig.l.) 
For very wide pulleys 
it is always better to use 
two sets of arms. For 
small pulleys, under 12 
inches in diameter, 4 
arms are better than 6, 
as they are less liable to 
break while being cast. 
Using 4 arms, the width 
of the arm at /i, in a 
pulley 10 inches in di- 
ameter, may be lj\ in. ; 
pulley 8 inches in di- 
ameter, lyi in. ; pulley 6 
inches in diameter, 1 
inch ; and the thickness 
and taper as given 
above. When pulley arms crack from shrinkage in casting, the 
trouble may usually be prevented by either increasing the thick- 
ness of the rim of the pattern or by reducing the size of the hub, 




PULLEYS. 349 

or both ; it will also help the matter to remove the core and the 
sand from the hub as soon as possible after the pulley is cast, 
and leave the casting in the sand undisturbed until cool. 

When the diameter of the shaft is less than 4 inches, the di- 
ameter of the hub is usually made twice the diameter of the 
shaft. When shafts are over 4 inches in diameter the hub of 
the pulley is usually made a little less than twice the diameter 
of the shaft. The length of the hub may be made three-fourths 
the width of the rim, for a tight pulley, and five-fourths the width 
of the rim for a loose pulley. 

The thickness of rim, measured at the edge, is usually : 

For pulleys under 12 inches in diameter, T \ inch. 

For pulleys from 12 to 24 inches in diameter, %. inch. 

For pulleys from 24 to 36 inches in diameter, T % inch. 

For pulleys from 36 to 48 inches in diameter, T % inch. 

For pulleys from 48 to 60 inches in diameter, l / 2 inch. 

For double belts increase the thickness of the rim one- 
eighth of an inch. 

The thickness in the middle may be about 1%. times the 
thickness at the edge. 

Pulleys which are to run at high velocity ought to be turned 
both inside and outside, in order to be in good balance. Pulleys 
to go on line shafts ought to be made in halves, so that they can 
be put on and taken off the shaft with convenience. Pulleys on 
which the belts are to be shifted must be a little over twice as 
wide as the belt, and they should be turned straight across the 
face on the outside. Pulleys on which the belts are not to be 
shifted ought to be only 1.2 times as wide as the belt, and they 
ought to be turned curved across the face; that is, the outside 
diameter of the pulley must be largest at the middle. Most fre- 
quently a straight taper is turned each way from the middle to 
the edges, and the following proportions will give good results : 

Pulleys under six inches wide, %-inch taper per foot. 

Pulleys from 6 to 12 inches wide, j^-inch taper per foot. 

Pulleys from 12 to 18 inches wide, ^-inch taper per foot. 

When pulleys are turned in a lathe where the tail-stock can be 
set over, a taper of ^-inch per foot is practically obtained when 
the tail-stock is set over ^g-inch per 1 inch length of arbor. For 
instance, if a crown pulley is to be turned ^-inch per foot, and 
the arbor is 12 inches long, the back center must be set over \\ 
= 2^-inch. If the arbor had been 14 inches long the back cen- 
ter would have had to be set over |f = T Vinch to obtain the 
same result. 

All pulleys must be well rounded on the edges. They must 
also be carefully balanced, especially if they are to run at high 
speed. 

Loose pulleys ought to have longer hubs than tight pulleys. 
They ought never to have hubs shorter than the width of the 
rim, and must always be provided with means for oiling. 



35 O PULLEYS. 

Stepped Pulleys. 

Stepped pulleys, or cone pulleys, as they are usually called, 
may be considered as several pulleys of different diameters cast 
together. Their proportions and sizes are calculated to get the 
required changes of speed, and the belt must have practically 
the same tension on all the different changes. 

Frequently it is required to have both pulleys of the same size, 
in order that they may be cast from the same pattern. In such 
cases the shaft of constant speed (usually a counter-shaft) must 
be run at a velocity equal to the square root of the product 
of the fastest and the slowest speed of the shaft of changeable 
speed (which usually is a spindle in a lathe or a similar machine). 
For convenience, in the following formulas we will call the 
driver, which is the shaft of constant speed, a counter-shaft, and 
the shaft of changeable speed, a spi7idle. 

The number of revolutions of the counter-shaft per minute 
is calculated by the formula : 

N—s/ F X S 

N = Number of revolutions of the counter-shaft per 

minute. 
F — Number of revolutions of the spindle per minute, 

when run at its fastest speed. 
S ■= Number of revolutions of the spindle per minute, when 

run at its slowest speed. 
The diameter of either the largest or the smallest step is 
then obtained by choosing one diameter and calculating the 
other by the formula : 

D — d X N d - D X n 



N 

D — Diameter of largest step on spindle. 

d = Diameter of smallest step on counter-shaft. 

n = Slowest number of revolutions of the spindle per 
minute. 

N = Revolutions of the counter-shaft per minute. 

The intermediate steps may be obtained by drawing a 
straight line, a b, and constructing steps within the angle 
formed by the line a b and the center line (see Fig. 2). The sum 
of the diameters of the two opposite steps will then be equal, 
and this is the way in which stepped pulleys may primarily be laid 
out whether both pulleys are of the same size or not. After- 
wards the diameters will have to be slightly changed, in order 
to give the belt the same tension on any of the different steps, 
as explained further on. 

Example 1. 

A pair of stepped pulleys, for four changes of speed, both 
pulleys of the same sige, are to be used on a milling machine 
spindle and its counter, the fastest speed to be 250 revolutions. 



9 


150 X 13 


11 


150 X 11 


13 


150 X 9 



PULLEYS. 351 

and the slowest speed 90 revolutions, per minute. The diam- 
eter of the largest step is 15 inches. What should be the speed 
of the counter-shaft, and what is the diameter of each inter- 
mediate step ? 

Solution : 

Speed of counter = s/ 90 X 250 = 150 revolutions per 
minute. 

The diameter of the largest step is 15". 

90 x 15 
Diameter of smallest step = — prx — = 9 inches. 

By the method as shown in Fig. 2, the intermediate diam- 
eters are. found to be 11" and 18". The speed of spindle will be : 

150 X 15 

First speed = = 250 revolutions per minute. 



Second speed = ^j = 177 revolutions per minute. 

Third speed = — — 127 revolutions per minute. 



Fourth speed = — = 90 revolutions per minute. 

15 

These calculations are only correct for speed, and must be 
slightly modified in order to get the proper tension on the belt, 
if an open belt is used ; for a crossed belt the tension is cor- 
rect if the pulleys are laid out in this manner. (See page 352.) 

When the number of revolutions per minute for each change 
of speed is given, the diameters of the intermediate steps ma}', 
with regard to speed, be calculated by the iollowing formulas : 

n {D + d)X N 
Vl - n +JV 
Di= Diameter of any step on spindle. 
D = Diameter of largest step on spindle. 
d — Diameter of smallest step on counter-shaft. 
11 — Revolutions of spindle per minute, corresponding to 

the diameter D\. 
N = Revolutions of counter-shaft per minute. 

After the diameter of any step on the spindle is calculated, 
-the diameter of the corresponding step on the counter-shaft may 
be obtained by subtracting the diameter of the step on the spin- 
dle from the value of (D + d). 

Example 2. 

A lathe spindle is required to run at 40, 120 and 300 revolu- 
tions per minute, and the diameter of the largest step is \6 
inches. Calculate speed of counter-shaft and diameter of steps. 



352 



PULLEYS. 



Solution : 

Speed of counter-shaft will be: 

N — V F X ^ = V 360 X 40 = 120 revolutions per minute. 
Diameter of smallest step on spindle will be : 



d = 



I S X 4 
120 



= 6 inches. 



Diameter of the intermediate step on spindle will be : 
= (18 + 6) X 120 = 
(120 + 120) 
Thus, the sizes of each step, with regard to speed, should 
be 6, 12 and 18 inches, but with regard to belt tension these 
sizes have to be slightly altered. 



To Correct the Diameter of Stepped Pulleys so that the 
Belt will have the Same Tension on all the Steps. 

At first thought, it may seem as if the belt would have 
equal tension on each step when the sum of the diameters of 
the largest and the smallest steps of the two pulleys are equal 
to the sum of the diameters of the two middle steps : but this is 
only correct if a crossed belt is used on the pulleys. For a two- 
step pulley it is also correct for either open or crossed belt if 
both pulleys are of the same size ; but if the pulleys are of dif- 
ferent sizes, the diameter of the steps must be calculated for 
two steps as well as if there were more. 

It is evident from Fig. 2, that an open belt will be tighter 
over the largest and the smallest pulleys than it would be over 
the two middle pulleys, as the part a of the belt runs parallel to 
the center line and will be as long as the distance between 
centers, but the inclined line, b, will be as much larger as the 
distance dto e. (See Fig. 2). 




A convenient way to solve this is : First calculate pulleys 
that will give the required speed, and of such sizes that the sum 
of the diameters of the two steps which are to work together 
will be equal, then calculate the length of the belt when laying 
on the largest and smallest steps, with a given distance between 



PULLEYS. 353 

the centers of the shafts. Then, by calculating the same way, 
try the belt on the other steps, which will then have to be cor- 
rected until the belt will fit each of the different pairs of steps. 

The length of the belt can be most conveniently calculated by 
the geometrical rule that the square of the perpendicular added 
to the square of the base is equal to the square of the hypothe- 
nuse. (See page 150.) The space between the centers of the 
shafts is considered as the base, and the difference in radius of 
the two corresponding steps is considered as the fte?-fte7idicular, 
which are both known, and from this the length of the line b is 
calculated (see Fig. 2), which is considered as the hypothenuse. 
Assuming that the belt covers half the circumference of both 
pulleys, the length of the belt can be found by adding half 
the circumference of each step to twice the length of b. 

Note. — This mode of calculation is not exactiy correct, but 
is very well within practical requirements. 

The length of half the circumference of the pulley is most 
conveniently obtained by the use of Table No. 24, page 209, by 
dividing the circumference of the corresponding circle by 2. 

A practical rule is simply to calculate the distance from d 
to e, and for each J^-inch the belt is found to be too ^long, add 
J^-inch to the diameter of the corresponding step on each pulley. 

For instance, the stepped pulleys in Example No. 2 are cal- 
culated so that they will give the required speed to the machinery 
when the three steps are 18, 12 and 6 inches in diameter and 
both pulleys are equal. Assume the distance between centers 
to be 5 feet. What will be the diameter of the middle step, after 
it has been corrected so that it will give the right tension to the 
belt? 

Solution : 

Five feet = 60 inches, and the difference between the 
radius of the corresponding steps is 9 — 3 = 6 inches. The 
distance from e to d will be : 

x — V60 2 + 6 2 — 60 = V3636 — 60 = 60.3 — 60 = 0.3 

Thus, each part of the belt will be 0.3" too long, or the whole 
belt will be 0.6" too long when on the middle step ; therefore, in 
order to make up for this, the middle step on each pulley must 
be increased gV = T y in diameter. Thus, the middle step on 
each pulley will be 12 T \ inches instead of 12 inches in diameter ; 
but, as both pulleys are increased, this does not change the 
relative speed of the shafts when the belt is on the middle step, 
and the similarity of the pulleys is also preserved, which will 
admit that both may be cast from the same pattern. 

The square root of 3636 may be obtained by use of log- 
arithms (see page 71), thus : 

, log. 3636 3.560624 

Log. V 3636 = -Sj = g = 1.780312 

and the number corresponding to this logarithm is 60.3. 



354 PULLEYS. 

Stepped Pulleys for Back=Qeared Lathes. 

On machinery havinc; :':::Ut::^ reducinc; pearinc;. such as 
lathes. milllmp machines, e:: . :: is drecuenzly me aim :: the 
l7;imiz-r z: arrange the speed :: the czunter and the diameters 
g: thle didz'erent steps ::' the zzne p alley in such przpzrzizns mat 
the sane rati: :: speed vrhll z e maintained zn eazh ^zep ii: als : 
from the si: est sired, v-ith hack pears cvr. :: the fastest 
speed. " dth hack pears :>:. ".V;.ti the rariz :z the hack pearmp 
is given, the ratio of speed for each step will be obtained by the 
formula: 

S = Rati: ::' speed izr eazh step. 

; y = Xumlzer ;■: zhanpes :z speed on die cone p alley 

i? = Reduction of speed by the back gearing. 

Tz:.-.::?lz. 

zzne pulley has ■: zhanpes :: speed. The larpest diameter :: 
czne pulley zn uhe spindle is 10 : j inches The zzne pulley :n 
the : : mrer-shait is :: he zz :he same size is nze zzne p alley zn 
the spmzile, md m even n:i: zz speed :s :: he maintained 
thronphcuz zize iz:lt ranpe zi the ten zhanpes :z sired. 
The slowest speed, —hen hack pears ire in. is i revzluvlzns per. 
zzite lalznlaze the speed zz nze zzunter-shadu the speed :i 
the spindle zzr eizh zizmpe. md tne dumezer zz eizn step zn 
the cone pulley of the spindle. 
Szlutizn: 
The ratio of speed for each step will be : 

: /— •-"*" " = - : ' =;.!■> - 

\ - \ 5 

The zzrresp indinp nnnzzer is l.-'l'i'. 

With back gears in. the speed of spindle : 

On first cone is 6 revolutions per minute. 

On second cone is 6 X 1.516 = 9 revolutions per minute. 

On third cone is 9 X 1151(5 = 14 revolutions per minute. 

On fourth cone is 14 X 1.516 = 21 revolutions per minute. 

On fifth cone is 21 X 11516 = 32 revolutions per minute. 

With back gears out, speed of spindle: 
On first cone will be 6X8= 48 revolutions per minute. 
On second cone will be 9X8= 72 revolutions per minute. 
On third cone will be 14 X 8 = 112 revolutions per minute. 
On fourth cone will be 21 X8 = 168 revolutions per minute. 
On fifth cone will be 32 X 8 = 256 revolutions per minute. 



PULLEYS. 355 

The speed of the counter-shaft will be : 

iV= V^ 4S X 256 = 112 revolutions per minute. 

As the speed of main lines in factories usually runs at some 
multiple of 10, we may, for convenience in getting even-sized 
pulleys for connections between counter and main shaft, in 
practical work, decide to run the counter-shaft 110 revolutions 
per minute. 

(When a pair of cone pulleys has an uneven number of 
steps, and are cast from the same pattern, the speed of the 
counter should be equal to the speed of the machine when the 
belt is run on the middle step). 

The diameter of the largest step of the cone pulley on the 

spindle is 10* inches. The corresponding step on the counter 

10 V X 48 
will be — vtq — = 4.581"; practically, 4*" diameter. 

The largest and smallest step on the counter-shaft will also 
be 10* and 4* inches in diameter. 

Any of the intermediate steps on the spindle may be cal- 
culated by the formula : 

D (D + d)Y.N 
n + N 

Di = (10* + 4*) X 110 = 9065 ticall 9 in . 

72 4-110 '* J 

D>= < 10 * + Y 2) X n ° = W inches. 
110 + 110 7 

Ds = qo* + 4/ 2 ) x no = 5932 practicall 6 in . 

168 -f 110 ' F J ' 

Thus, assuming the counter-shaft to run 110 revolutions per 
minute, the speed of the spindle, with back gears out, on the five 
different steps will be : 

— ^ = 265 revolutions per minute. 



110 X 9 
6 



= 165 revolutions per minute. 



110 X IV 

?=- = 110 revolutions per minute. 



110 X 6 
9 

no x 4y 2 
ioy 2 



= 73 revolutions per minute. 
= 47 revolutions per minute. 



356 PULLEYS. 

When the back gears are in action the speed will be : 

: — = 33 y^ revolutions per minute. 
8 

165 

-— = 20^ revolutions per minute. 

= \Z% revolutions per minute. 

8 

73 

_L_ = 91^ revolutions per minute. 

8 

47 
. — : = 5^ revolutions per minute. 

8 

These speeds are all within the practical requirements of 
the problem, and now the next operation is to modify the diam- 
eters slightly in order to get proper tension on the belt. (See 
page 352.) 



FLY=WHEELS. 



Fly-wheels are used to regulate the motion in machinery by 
storing up energy during increasing velocity, and giving out 
energy during decreasing velocity. Fly-wheels cannot perform 
either of these functions without a corresponding change in 
velocity. The rim of the wheel may be very heavy and moving 
at a high velocity, the change in speed may be small and hardly 
perceptible if the energy absorbed and given out is small, but 
there must always be a change in velocity to enable a fly-wheel 
to act. The common expression of gaming power by a heavy 
fly-wheel is very misleading, to say the least. There is no 
power gained by a fly-wheel but, on the contrary, considerable 
power is absorbed by friction in the bearings when a shaft is 
loaded with a heavy fly-wheel, (see example in calculating fric- 
tion, page 305). Nevertheless, a fly-wheel performs a very use- 
ful function in machinery by storing up energy when the supply 
exceeds the demand and giving it out at the time it is needed to 
do the work. ( For momentum of fly-wheels see example, page 
300. For kinetic energy, see example, page 301 ). 

Weight of Rim of a Fly=Wheel. 

The weight of a rim of a cast-iron fly-wheel will be : 
W—d 2 X 0.7854 X D X 3.1416 X 0.26 ; this reduces to, 
W—D X d 2 X 0.64 
D = Middle diameter of rim in inches. 
d= Diameter of section of rim in inches. 
W = Weight in pounds. 



FLY-WHEELS. 357 

Example. 

A round rim of a fly-wheel is 4 inches in diameter and the 
middle diameter of the wheel is 36 inches. What is the weight 
of the rim ? 

Solution : 

Jf=36X4X4X 0.64 = 369 pounds. 

For a rim of rectangular section the weight will be : 
W = Width X thickness X D X 3.1416 X 0.26 
IV = Width X thickness X D X 0.816 

Example. 

The width of the rim is six inches, the thickness is two 
inches, and the middle diameter of the rim is 4S inches. What 
is the weight of the rim ? 

Solution : 

^=2X6X48X 0.816 = 470 pounds. 

Centrifugal Force in Fly=Wheels and Pulleys. 

Pulleys are not only liable to be broken by the stress due to 
the action of the driving belt, but in fast-running pulleys and 
fly-wheels the stress due to centrifugal force is far more dan- 
gerous. This stress increases as the square of the velocity and 
directiy as the weight, therefore there is a limit to the velocity 
at which fly-wheels and pulleys can be run with safety. 

Generally speaking, increasing the thickness of the rim 
does not increase its strength, because the total tensile strength, 
the total weight of the rim, and. consequently, also the 
centrifugal force, increase in the same proportion : but it has 
great influence upon the strength of the wheel to have the ma- 
terial in the rim distributed to the best advantage. At the same 
time it is very important to construct the rim and arms of such 
proportions that the initial stress due to uneven cooling in the 
foundry, is avoided. 

The common formula is : 

„ ■ .., , , Mass X (velocity) 2 

Centrifugal force = ^ — 

radius 

m _ Weight T7 . . n X r X2tt 

MaSS = -32"2- Velocity = - —^- 

Therefore, 



cf = 



cf = 



. * {^^-y 



32.2 X r 
WX n 2 X r 2 X 0.01096628 



32.2 X r 
cf — W X « 2 X rX 0.00034 
cf = Centrifugal force in pounds. 



358 



FLY-WHEELS. 



W = Weight of revolving body in pounds. 
n = Number of revolutions per minute. 
r ■=■ Middle radius of pulley rim in feet. 
Thus, for any body whose ce?iter of gravity swings in a 
circle of one foot radius, at a speed of one revolution per min- 
ute, the c e?itr if ugal force will be 0.00034 times the weight of 
the body. 

Example. 

The rim of a fly-wheel is five feet in middle radius and 
weighs 8000 pounds. It makes 75 revolutions per minute. 
What is the stress due to centrifugal force? 

Solution : 
Centrifugal force = 8000 X 75 2 X 5 X 0.00034 = 76500 pounds. 

This is the total centrifugal force 
tending to burst the rim, ( see arrows in 
Fig. 3 ) ; the force tending to tear the 
rim asunder in anv two opposite points 

l ■ 76500 10-!" A 

as a, b, is . , _, t _, a ^ = 121 10 pounds. 



FIG 3. 




_1416 X 2 

The next question is : Has the sec- 
tion of the rim tensile strength enough to 
resist this stress with safety? If not, 
either decrease the rim speed or make the rim of material having 
more tensile strength. 

The centrifugal force for the same number of revolutions 
increases as the radius, therefore the average centrifugal force 
acting in the arms is only about half of the centrifugal force 
acting in the rim, and as the stretch is in proportion to the stress, 
the rim tends to stretch more than the arms, and. consequently, 
it can not vield freelv to the action of the centrifugal force, 
but is to a 'certain extent held back at the junction with the 
arms. This action is shown in an exaggerated form at a, Fig. 4. 
In regard to this action, the part of the rim between the 

arms may be considered 
as a beam fastened at both 
ends and uniformly loaded 
throughout its whole length 
equal to that amount of cen- 
trifugal force in the rim 
which is resisted by the 
arms : therefore, the rims of 
large pulleys should always 
be ribbed on the inside. 
(See cross-section of rim at 
X, Fig. 4). 

Another bad feature frequently seen in pulleys is the 
counter-balance. (See b. Fig. 4). This little piece itself, weigh- 
ing probably only five pounds, holds the pulley neatly in balance 





FLY-WHEELS. 359 

and is very innocent as long as the pulley is standing still, but 
imagine what stress it will produce on the rim of a 6-foot pulley 
running at a rim speed of 80 feet per second. 
Solution : 

cf= 8 ° 2 X -A = 331 pounds. 
3 X 32.2 

Thus, when that pulley is running at a speed of 80 feet per 
second this counter-balance of five pounds will produce the 
same stress as if it was loaded with 331 pounds when standing 
still; therefore, it is evident how important it is to turn fast- 
running pulleys both inside and outside in order to reduce 
counter-balancing to the least possible amount. 

The danger of the rim deflecting or breaking from the stress 
due to the resistance from the arms (as shown in Fig. 4), can be 
avoided by running ribs on the inside of the rim, and the danger 
caused by counter-balancing can be entirely eliminated by mak- 
ing the pulley balance without adding any balancing pieces. 
Thus, one danger of breaking is avoided by proper designing 
and the other by good workmanship. 

The direct action of the centrifugal force on the rim is cal- 
culated by the formula, 

cf= w X n 2 X r X, 0.0)034, 
and the weight of the rim of a cast-iron fly-wheel having one 
square inch of sectional area and a radius of one foot will be 
2 X 77 X 12 X 0.26 pounds, and a ring of r-foot radius will 
weigh ;- X 2 X 77 X 12 X 0.26 pounds. As already stated, the 
centrifugal force increases as the square of the velocity ; that 
is, if the number of revolutions is doubled the centrifugal force 
is increased four times ; thus is the dangerous limit approached 
very rapidly under increased speed, and in order to prevent 
accident, if the speed should happen to increase, it is necessary 
always to use a high factor of safety in such calculations. 
Thus, using 15 as a factor of safety* and assuming the tensile 
strength of cast-iron as 12,000t pounds per square inch, the 
stress in each cross-section at a and b must not exceed 800 
pounds per square inch. The allowable centrifugal force in 
both sections of the rim is 2 X 800 pounds, and inserting those 
values and solving for n the greatest number of revolutions 
allowable for a cast-iron fly-wheel will be : 

2 77 X 12 X 0.26 X r X » 2 X r X 0.000340568 = 800 X 2 
n -2 r 2 = 752891 

ur= VT52801 

_ 868 = 868 

r n 

* 15 as factor of safety with regard to strength, is only \ 15 = 3.873, or 
less thnn 4, as factor of safety with regard to speed. 

t This tensile strength for cast-iron may seem very low, but it is dangerous to 
assume more, because of inside stress in arms or rim already, due to uneven cool- 
ing of the casting in the foundry. 



360 FLY-WHEELS. 

Transposing this to diameter in inches, the constant will be 
24 X 868 = 20832. The formula will be : 

9QOOQ 

Number of revolutions per minute = 



Diameter in inches 



Diameter in inches. 
20832 



Revolutions per minute. 



Rule for Calculating Safe Speed. 

Divide 20832 by the diameter of the fly-wheel in inches, and 
the quotient is the allowable number of revolutions per minute 
at which a well-constructed fly-wheel may be run with safety. 

Rule for Calculating Safe Diameter. 

Divide 20832 by the number of revolutions per minute, and 
the quotient is the safe diameter in inches for a well-constructed 
fly-wheel. 



SHAFTING. 

When calculating strength of shafting, both transverse and 
torsional stress should be considered. Transverse stress is 
produced by the weight of the shaft itself, the pulleys and the 
tension of the belts, the effect of which is very severe if the 
distance between the hangers is too long. Torsional stress is 
produced by the power which the shaft transmits. Usually the 
distance between the hangers is made so short that the torsional 
stress on a shaft is the greater. For transverse stress the shaft 
may be considered as a round beam, supported under the ends and 
loaded somewhere between supports. According to Table No. 
30 the transverse stress which will destroy a wrought iron beam 
one inch square, fastened at one end and loaded at the other, is 
600 pounds ; the strength of a round beam of the same diameter 
is (see page 251) 0.6 that of a square beam. When the beam 
is supported under both ends and loaded in the middle, its 
breaking load will increase four times ; therefore, the constant, 
c, will be 600 X 4 X0.6 = 1440. Using 10 as factor of safety, 
the formula for transverse strength of a wrought iron shaft will 
be: 

D = Diameter of shaft in inches. 
L = Distance between hangers in feet. 
W = Transverse load in pounds, supposed to be at the 

middle, between the hangers. 
144 = Constant for wrought iron, and 100 to 120 may be 
used as constant for cast-iron, with 10 as factor of safety for 
transverse strength^ 



SHAFTING. 361 

Formula 1, expressed as a rule, will be: 

Multiply the distance between hangers, measured in feet, 
by the transverse load in pounds ; divide this product by 144, and 
the cube root of the quotient will be the diameter of the shaft 
in inches, calculated with 10 as factor of safety for transverse 
strength. 

Shaft not Loaded at the Middle Between the Hangers. 

When a shaft is not loaded at the middle of the span, but 
somewhere toward one of the hangers, it will carry a heavier 
load, with the same degree of safety, than it would if loaded in 
the middle, and the ratio is in inverse proportion as the square 
of half the distance between hangers to the product of the short 
and the long ends of the shaft. For instance, a shaft is six feet 
between hangers and loaded at the middle. What would be the 
difference in transverse strength if it was loaded two feet from 
one hanger and four feet from the other ? 

3 X 3 = 9 and 2 X 4 = 8. 

Thus, find the transverse load for a shaft when loaded in 
the middle, multiply by 9 and divide by 8, and the quotient is 
the load which the same shaft will carry with the same degree 
of safety against transverse stress, if loaded two feet from one 
end and four feet from the other. 

This rule only applies to the transverse strength, and not to 
the transverse stiffness of the shaft. For different shapes of 
shafts and different modes of loading, see beams, pages 243-244. 
When shafts are heavily loaded near one hanger, and the hanger 
on the other side of the pulley is further off, most of the load is 
thrown on the bearing nearest to the pulley, and this bearing is, 
therefore, liable to heat and to cause trouble, even if the shaft 
is both stiff and strong enough. ( See reaction on the support 
of beams, page 252). 

Transverse Deflection in Shafts. 

The transverse deflection in a shaft may be calculated by 
the formula : 

4 3 



V=^jL 1 JVC_ L = ^ 



S D± 



s > w c 

w- SD * s- L * wc 

S = Deflection in inches. 
D = Diameter of shaft in inches. 

L = Length of span in feet. 
W= Load on middle of shaft in pounds. 

C= Constant = 1.7 X constant in Table No. 31, and for 
wrought iron or Bessemer steel may be taken as 0.00002652, 



362 SHAFTING. 

Constant C may be calculated from experiments by the 

formula. 

Cr- SI?i 



L ? - W 

S= Deflection in inches noted in the specimen, when 
supported under both ends and loaded transversely 
at the middle between supports. 

D = Diameter of specimen in inches. 

L = Distance between supports of specimen in feet 
?F= Experimental load in pounds. 

Example. 

A round specimen placed in a testing machine, supported 
under both ends and loaded at the middle with 2000 pounds, 
deflects 0.1 inch. The diameter of the specimen is two inches 
and the distance between supports is three feet. Calculate 
constant C for this kind of material. 

Solution : 

0.1 X 2 4 



C 



3 s X 2000 



C — 1S = 0.0000296 inch. 
54000 

Thus, the deflection for this kind of material is 0.0000296 
inch per pound of load, applied at the middle, between supports, 
for a round bar one inch in diameter and one foot between 
supports. 

Allowable Deflection in Shafts. 

The distance between the hangers must always be deter- 
mined with due consideration to the allowable transverse deflec- 
tion in the shafting, especially when the shaft is loaded with 
large pulleys and heavy belts, remembering that the deflection 
increases directly with the transverse load and with the cube of 
the length between the bearings, (see page 254). The allowable 
transverse deflection in shafting ought not to exceed 0.006 to 
0.O0S inch per foot of span (see page 266;. A beam of wrought 
iron one foot long and one inch square, when supported under 
both ends and loaded at the middle, will deflect 0.0000156 inch 
per pound of load, (see Table No. 31. page 259;. and a round beam, 
deflects 1.7 times as much as a square beam, when the diameter 
and side are equal. A round shaft, one inch in diameter and 
one foot long, when loaded at the middle with 144 pounds will, 
therefore, deflect 144 X 1.7 X 0.0000156 = 0.003S2 inch. 

Thus, this load does not give more than an allowable de- 
flection. But. suppose the distance between bearings is doubled 
and the load decreased one-haif : the ultimate strength of the 
shaft will be the same, but the deflection will be 72 X 1.7 X 2 3 X 
0.0000156 == 0,01528 == 0.0764 inch per foot, 



SHAFTING. $6$ 

This calculation shows plainly how very necessary it is to 
have bearings near the pulleys where shafts are loaded with 
heavy pulleys and large belts. There is nothing more liable to 
destroy a shaft than too much deflection, because the shaft is, 
when running, continually bent back and forth, and at last it 
must break. The fact must never be lost sight of that strength 
and stiffness are two entirely different things and follow entirely 
different laws; therefore, after calculations are made for 
strength, the stiffness must also be investigated, as stiffness is 
a very important property 7 in shafting. The best way to over- 
come too much transverse deflection is to shorten the distance 
between the bearings. Of course, increasing the diameter of the 
shaft will also overcome deflection, but shafting should never be 
larger in diameter than necessary, because the first cost increases 
with the weight, which increases as the square of the diameter, 
and the frictional resistance will also increase with the increased 
diameter ; consequently, also, the running expenses. 

Torsional Strength of Shafting. 

Shafting may be considered as a beam fastened at one end 
and having a torsional load applied at the other end equal to 
the pull of the belt on an arm of the same length as the radius of 
the pulley. In Table No. 32, page 268, constant c is given as 580 
pounds for wrought iron. 

The formula for twisting stress, as explained under beams 
( see page 267) is, 

p = D%c d = SZJL 

m \ c 

in = the length of the lever or arm in feet, and will here 
be the radius of the pulley and be denoted by r. The length of 
the shaft has no influence on its torsional strength, but only on 
its angle of torsional deflection (see page 268). Using 10 as 
factor of safety, the formula will be : 

D =XrW 



^ 



58 

D = Diameter of shaft in inches. 
r — Radius of pulley in feet. 

W ' = Pull of belt in pounds. 

58 — Constant, with 10 as factor of safety = Vio X 580, 
taken from Table No. 32, page 268. 

Frequently it is mere convenient to calculate the torsional 
strength of shafting according to the number of horse-power the 
shaft is to transmit (see page 317). 

In the above formula, assume W to be 58 pounds, r to be 
one foot, and D will be one inch. That is, a shaft one inch in 
diameter is strong enough \q resist with 1Q as factor of safety, 



364 SHAFTING. 

a torsional load of 58 pounds acting on an arm one foot long. 
Assuming this 58 pounds to act on the rim of a pulley of one 
foot radius, two feet in diameter, and making one revolution per 
minute, it will transmit power at a rate of 58 X 6f = 364f foot- 
pounds per minute ; but one horse-power is 33,000 foot-pounds 
per minute, and if the shaft should transmit one horse-power it 

, 33000 , . . „ ■ , 

must make - oaAi — 90.52 revolutions per minute. Hence the 

practical formulas for torsional strength of shafting : 

D * HlXW H=P**» n= I/X ^° 

\ J t 90 D* 

D = Diameter of shaft in inches. 

H = Number of horse-power transmitted by the shaft. 
n = Number of revolutions made by the shaft per minute. 
90 = Constant, using 10 as factor of safety, and assuming 
the torsional strength to be as given in Table No. 32. 



Torsional Deflection in Shafting. 

In constructing different kinds of machinery it is frequently 
necessary to consider the torsional deflection. The formula 
for torsional deflection for wrought iron ( see page 271) will be : 

c _ 0.00914 X mLP 

jyi This will transpose to 

s _ 48 H L 
~TiD± 

S= Deflection in degrees. 
H = Number of horse-power transmitted. 

0.00914 X 33000 _ , ft 
48 = Constant ; calculated thus, — 2X3 1416 — ' — 

L = Length of shaft in feet between the force and the 

resistance. 
n = Number of revolutions made by the shaft per minute. 
D = Diameter of shaft in inches. 

Example. 

How many degrees is the deflection of a shaft two inches 
in diameter, 50 feet long, making 300 revolutions per minute and 
transmitting 15 horse-power, applied at one end and taken off at 
the other ? 

Solution : 

48 HL 48X15X50 ,_ . , 

6 = ~^D^ = - 300X2* - = 7 ^ dGgreeS - 



SHAFTING. 



365 



Classification of Shafting. 

Shafting may be divided into three different kinds. 

First.— Shafts where the main belts are transmitting the 
power, or so-called " Jack Shafts.'* Such shafts must have then- 
boxes as near the pulleys as possible. For torsional strength 
their diameter may be calculated by the formula, 



D 



=v 



H X 125 



n 



(See Table No. 40.) 



Second. — Common shafting in shops and factories, where 
the power is taken off at different places for driving machinery. 
Such shafts ought to be supported by hangers as given in 
Table No. 43, and their supports must also be reinforced by 
extra hangers, if necessary, where an extraordinary large pulley 
or heavy belt is carried. For torsional strength the diameter of 
such shafts may be calculated by the formula, 



D = J&X90 



\ 



(See Table No. 41.) 



Third. — Shafting having practically no transverse stress, 

but used simply to transmit power from one place to another. 

Such shafts ought to be supported by hangers according to 

Table No. 43, and the diameter may be calculated by the 

formula, 

3 



D — 



\ 



H X 50 



(See Table No. 42.) 



TABLE No. 40.— Giving Horse=Power of Main Shafting 
at Various Speeds. 



Diameter 
of Shaft 
in Inches. 


Revolutions per Minute. 


60 

2.6 


SO 
3.4 


100 


125 150 


175 


200 225 


250 


275 


1 300 


1# 


4.3 


5.4 6.4 


7.5 


8.6 9.7 


10.7 


11.8 


12.9 


2 


3.8 


5.1 


6.4 


8 


9.6 


11.2 


12,8 


14.4 


16 


17.6 


19 


2* 


5.4 


7.3 


9.1 


11 


13 


16 


18 


21 


23 


25 


27 


2% 


7.5 


10 


12.5 


15 


IS 


22 


25 


28 31 


34 


37 


2H 


10 


13 


16 


21 


25 


29 


33 


37 42 


46 


50 


3 


13 


17 


21 


27 32 


38 


43 


49 


54 


59 


65 


3* 


16 


22 


27 


34 41 


4S 


55 


62 


69 


76 


82 


%% 


20 


27 


34 


43 51 


60 


6S 


77 86 


94 


103 


3& 


25 


34 


42 


53 I 63 


74 


84 


95 105 116 


126 


4 


30 


41 


51 


64 77 


90 


102 


115 128 141 


154 


4# 


43 


58 


73 


91 109 


128 


146 


164 182 201 


219 


5 


60 


SO 


100 


125 ,150 


175 


200 


225 |250 |275 


•300 



366 



SHAFTING. 



TABLE No. 41.— Giving Horse=Power of Line Shafting 
at Various Speeds. 



'- 1 — 

3°j 


Revolutions per Minute. 


IOC 


125 
7.4 


ISO 


175 200 225 250 


275 300 


325 350 400 


1* 


6 


9 


10.4 


12 13.4 


15 


16.4 IS 


19.4 2l! 23.8 


1% 71 


9.1 


10.9 


12.7 


14.5 16 


18 


20 22 


23.8 25 29 


2 8.9 


11.1 


13 


15.5 17.7 20 


22 


24 27 


19 31 35 


2% 10.6 


13.2 


16 


18.5 21 


24 


27 


29 32 


34 37 42 


2# 12.6 


15.8 


19 


22 25 


28 


32 


35 38 


41 44 50 


2^ 15 


18.6 


22 


26 30 


33 


37 


41 44 


45 52 59 


'2% 17 


22 


26 


30 35 


39 


43 


48 52 


56 61 69 


2K 23 


29 


34 


40 46 


52 


58 


64 69 


75 81 92 


3 30 


37 


45 


52 60 


67 75 


BS 90 


97 105 120 


s% : - 


47 


57 


67 76 


S(> 95 105 114124 133152 


zy 2 4S 


59 


71 


83 95 


107 119 131 143155 167190 


3K 58 


73 


SS 


102 117 132 146 161 176190 205 234 


4 71 


89 


107 


125 142 160 17S 196 213 231 249 284 


TABLE No. 


42.— 


Giving Horse=Power of Shafting Used 




Only 


for Transmitting Power. 


*• — T 

- — B 

1 8-S 






Revolutions per Minute. 


Diam 

of S 

hi 1 ii« 


100 
: 6.7: 


125 


150 


175 200 


225 250 


275 


300 325 350 400 


IV? 


8.4 


10 


11.8 13.5 15.1 16.5 


18.5 


20.3 22 23 27 


in 


8.6 


10.7 


12.8 


15 17.1 


19.3 21.1 


23.6 


25.8 


28 30 34 


1H 


10.7 


13.4 


16 


18.7 21.5 


24 26.8 


29.4 


32.1 


35 37 43 


1% 


13.2 


16.5 


19.7 


23 26.4 


29.7 33 


36.2 


39.5 


43 46 52 


2 


16 


20 


24 


28 32 


36 40 


44 


48 


52 56 64 


2 Mi 


19 


24 


29 


33 


42 48 


53 


57 


62 67 76 


2* 


23 


28 


34 


40 45 


51 57 


63 


6S 


74 80 91 


2*j 


27 


33 


40 


47 54 


60 67 


74 


80 


87 94107 


2# 


31 


39 


47 


55 62 


70 78 


86 


94 


102 109 125 


2¥ 


41 


52 


62 


73 83 


93 104 


114 


125 


132 146 166 


3 


54 


67 


81 


94 108 


121 135 


14S 


162 


175 189 216 


3X 


69 


S6 


103 


120 137 


154 172 


189 


206 


223 240 275 


3>4 86 


::: 


128 


150 171 193 114 


236 


257 279 300 343 



SHAFTING. 



367 



Distance Between the Bearings. 

Jack shafts should always have bearings as near the pulleys 
as possible. 

Ordinary line shafts, as given in Table No. 41, and shafts 
for simply transmitting power, may have the distance between 
the hangers as given in the following table : 

TABLE No. 43. 



Diameter of Shaft in Inches. 


l l A to 1% 


2 to 2)4 


2y 2 to 4 


Distance between bearings in feet 


Qy 2 


8 


10 



Shafts for Idlers. 

Shafts for idlers ( see C, Fig. 1 ) have very little torsional 
stress and the distance between the bearings may also be very 
short, so that even with a great transverse load such a shaft 
may be of comparatively small diameter as far as requirements 
for strength is concerned. In such a shaft there is great danger 
of trouble from hot bearings ; therefore, in designing, it is very 
important to make its diameter and the length of the bearing of 
such proportions that excessive pressure per square inch of 
bearing surface is avoided. 

Example. 

Twenty-five horse-power is to be transmitted from A to B 
through idler C. ( See Fig. 1 ). 
are 36 inches in diameter and 
make 40 revolutions per minute 



The gears on shafts A and B 
fig. 



What is the necessary diameter 
of shaft C, which is supported 
by two bearings one foot apart 
and carrying a gear 48 inches 
in diameter placed at the mid- 
dle between the bearings. 

Solution : 

The velocity on pitch line of gear A will be 

40 X 36 X 3.1416 




= 377 feet per minute. 

33,000 X 25 = 825,000 foot-pounds. 



12 

25 horse-power 

The pressure at the pitch line of A transferred to Cwill be 
825000 



377 



2188 pounds. 



368 SHAFTING. 

The reaction at the pitch line between C and Z>, is also 
2188 pounds; therefore, the total pressure (besides the weight 
of C, which is omitted in this calculation ) on both bearings will 
be 2^ X 2188 = 4376 pounds and the pressure of each bearing of 
C will be 2188 pounds. Allowing a pressure on the bearings of 
100 pounds per square inch, the necessary bearing surface 
will be 

2188 

= 21.S8 square inches for each bearing. 



100 

Assuming the length of the bearing to be twice its diameter, 
D X 2£> = 21.S8 

j-)2 — 21.88 

2 

D =\/l0.94 

D = 3.3 inches. 

Calculating the size required with regard to transverse 
strength by the formula on page 360, 



D = JiiiiSTC = jnches> 

\ MA 



144 

Thus, a shaft 3.3 inches in diameter is of ample size for 
strength. The surface velocity of this shaft will be, 

3.3 X 3.1416 X 40 = „ 4 5 £eet 
12 

p,er minute, and at that velocity a pressure of 100 pounds per 
square inch of bearing surface is very safe from liability of 
heating if the bearing is well made and amply provided with oil. 



Proportion of Keys. 

The breadth of the key is usually made to be one-fourth of 
the diameter of the shaft, and the thickness to be one-sixth of 
the diameter of the shaft. 

Keys and key-ways are usually made straight and should 
always be a very good fit sidewise. Frequently set-screws are 
used on top of keys in mill gearing. Sometimes in heavy ma- 
chinery keys are made tapering in thickness, usually one-eighth 
inch per foot of length. A corresponding taper is made in the 
depth of the key-way in the hub. Key-ways in shafts are always 
made straight. 

For light and fine machinery taper keys are never used. 



SHAFTING. 



369 



TABLE No. 44.— Dimensions of Couplings for Shafts. 

(All dimensions in inches.) 





Dimensions 


of Couplings. 


Diameter of 


Number of 


Diameter 










Coupling 


Coupling 


of Shaft. 










d 


D 


Z 


t 


Bolts. 


Bolts. 


l* 


3 


7 


4 


H 


H 


4 


1# 


3^ 


8 


4^ 


X 


X 


4 


2 


4^ 


9 


4^: 


% 


% 


4 


2^ 


5 


10 


s# 


H 


% 


4 


3 


6 


11 j* 


6 


X 


% 


5 


sy 2 


Q7/S 


13 


6^ 


X 


% 


5 


4 


1% 


14^ 


IX 


1 


x 


6 


4X 


8X 


153* 


m 


IX 


1 


6 


5 


9 


17 


s% 


1% 


IX 


6 




BEARINGS. 

A satisfactory rule is to make the length of the bearing for 
line shafting six times the square root of the diameter of the 
shaft. 

Example. 

What is a suitable length of bearings for a shaft of four 
inches diameter? 

Solution : 

Length of bearing = 6 X \/ 4 = 12 inches. 

Some designers make the length of the bearing four times 
its diameter. 

Area of Bearing Surface. 

The projected area of any bearing is always considered as 
its bearing surface. Thus, the length of the bearing multiplied 
by the diameter of the shaft gives the area of bearing surface. 
For instance, the length of the box is twelve inches and the 
diameter, of the shaft is four inches ; the area of bearing surface 
is 12 X 4 = 48 square inches. 

Allowable Pressure in Bearings. 

The allowable pressure per square inch of bearing surface 
will depend on the surface speed of the shaft and the condition 



OJ 



BEARINGS. 



of the bearing, arrangements for oiling, etc. For common line 
shafting from two to four inches in diameter, not making over 
200 revolutions per minute, a pressure not exceeding forty 
pounds per square inch ought to work well. Greater pressure 
or greater speed may make it difficult to keep the bearings cool. 

Example. 

What pressure may be allowed on a bearing twelve inches 
long and four inches in diameter ? 

Solution : 

Pressure = 4 X 12 X 40 = 1920 pounds. 

In well constructed machinery there should not be any 
trouble from hearing, if the surface velocity and the pressure in 
the bearings does not exceed the values given in the following 
table :— 



Metric Measure. 



English Measure. 



Kilograms per Square, Surface Velocity in 
Meters per Minute. 



Pounds per Square Surface Velocity in 
Feet per Minute. 



_ ;~ :.— e:e: 



. r. : h 



D 

12 
20 



100 
20 



fa 
IS 

300 



300 

150 
60 



The bearings for machinery in general are constructed in 
various ways and of different proportions, according to the de- 
signer's judgment, but it is a well-known fact that high-speed 
machinery must have longer bearings than slow-speed ma- 
chinery. 

The length of the bearing will usually vary from one and 
one-half to six times the diameter. 

When the shafts are small (less than two inches in diameter), 
and the speed is from 100 to 1000 revolutions per minute, the 
following empirical formula may be used as a guide : 

L = length of bearing. d= diameter of bearing, n = num- 
ber of revolutions per minute. 

-b J FIG. 1 




BEARINGS. 



37* 



Figure 1 shows a cheap, solid cast-iron box used for com- 
paratively small and less important shafts. Dimensions, suita- 
ble for bearings from one to two inches in diameter, are given 
in the following table : — 









k All Dimensions ir 


Inches.) 








u 

<D 
-t-> 




T— 1 




St 

+ 

CO 


i— 1 
+ 




+ 


II 


11*8 




II 


II 


II 


II 


II 


II 


^ 


i-l 


*0 

g 


*& 


Vj 


^ 


v> 


■** 


§ 


d 


/ 


b 


C 


k 


e 


i 


m 


1 


1# 


iy 2 


13/ 


334 


5^ 


V* 


1# 


13* 


2* 


17/8 


2* 


4 ^ 


6^ 


9 
1 6 


9 

1 6 


itf 


i 1 / 


25/ 8 


^A 


25/ 8 


*tf 


</ 2 


5/8 


5/8 


I 1 / 


iK 


3 1 


2S/ 8 


»* 


6 


8'/ 


1 1 

1 6 


1 1 


13/ 


2 


3^ 


3 


3*/ 


63/ 


9^ 


V\ 


H 


2 



Figure 2 shows a babbitted split box suitable for shafts 
from one to four inches in diameter, and running at a com- 
paratively slow speed. 

d = Diameter of shaft. 
a = 2% X d. 

b = l% X d. 

c = 3 X d + K inch. 

k = 4X d+ 1% inches. 

e = % X d + % inch. 
f— % X d+ % inch. 

g = iy 2 x d. 

1=2 X d 

Thickness of babbitt metal, t = Vi6 d -\- l /z inch. 
Diameter of bolt, h = Vod + X inch. 
Diameter of bolt, i = Vkd -\- Y% inch. 

Figure 3 shows the same general design of box as Figure 
2, excepting that the bearing is longer and the base wider. 
This box is more suitable for comparatively high-speed shafts. 




372 



BEARINGS. 



FIG. 3. 



d = Diameter of shaft. 
a=l% d-\-l% inches. 
b = lftd + ft inch. 
c = 2%d+2 inches. 
k = 3 d + 3 inches. 
e = %d -f- % inch. 
/= %d + Xinch. 

/=5 xVd 
m= d 
h— % d+ % inch. 

i— yi d -\- Y% inch. 

Figure 4 shows a babbitted box or pedestal suitable for 
comparatively heavy-loaded shafts, from three to eight inches 
in diameter, such as outer bearings for steam engine shafts, 
bearings for jack shafts, etc. 



FIG. 4. 





d— Diameter of shaft; a = 2d -\- 1% inch; b = 1)4 d-\- ft 
inch; c = 2% d-\- 2 inches; k = 'd% d-\- 3 inches; e= %d+ % 
inch; f=%d-\- X m ch; g—l%d\ /=2d; m = d; 
D = lftd. 

Diameter of bolts, h = 0.2 d -\- %. inch (approximately). 

Diameter of bolts, i = 0.2 d 4- y% inch (approximately). 



BEARINGS. 



m 



Figure 5 shows a bearing 
fitted into the frame of a ma- 
chine, suitable for shafts from 
one to three inches in diameter. 
The cut shows a part of the 
head-stock of a speed lathe 
fitted with this kind of a bear- 
ing. The bearing itself, which 
may be of gun metal or cast 
iron, is carefully fitted into the 
frame by planing and scrap- 
ing. 

This kind of a bearing is 
sometimes lined with babbitt, 
but more frequently the spindle 
is carefully fitted into the bear- 
ing by scraping. 



FIG. 5. 




\JEEEE^L\ 



d = Diameter of bearing; l=2d-, a = 2%. d-\- 1 inch ; 
b — \y 2 d +% inch; c — \%d\ e = / — \%d + y 8 inch; 
f=e = l%d+y 8 mch; g=%d+ X inch ; /t = iy 8 d+y 8 
inch ; i == 2d. Diameter of screws = 3 Ae d -f- %6 inch. 

Figure 6 shows the form of a self-lining and self-oiling 
bearing, very suitable for high-speed machinery, and used to a 
great extent for dynamos and electric motors. The figure 
shows a part of a dynamo frame with the box in section, cut 
through the center line of bearing, and also a partly sectional 
cut from the top. For dimensions see Table No. 46. 

The bearing n n may be cast in one piece from gun metal, as 
shown in the cut, or it may be (preferably for the larger size) 
made in two parts. The seat for this box is turned in spherical 
form on the outside, and a fit is obtained between this bearing 
and the frame of the machine by casting in type metal or bab- 
bitt metal, as shown at m m. 

The loose rings, n 7i, are continually dipping into the 
oil reservoir, and carrying oil to the shaft. (Chains are fre- 
quently used instead of rings). The stop-rings should be set so 
that the spindle has room for a little motion lengthwise in the 
bearing. This will in a great measure prevent heating and cut- 
ting, and by their peculiar shape the stop-rings will, by the 
action of centrifugal force, throw the oil off at a a, to return to 
the oil reservoir ; h h are plugs in the oil hole ; the screw i pre- 
vents the box from turning with the shaft, and also forms a 
convenient projection to take hold of when taking the cap off 
of the bearing. 



%u 



EEARDfGS. 



Pis. 6. 




TABLE No. 


40 — Giving Dimensions of Fig. 


6. 


d D 


D\ 


L 


I 


h 


/ 


Inches. V&d- % 


l%d+% 


1W- 1 


4XV^ 


2\ d - 1 


Inch- 


1 1^ 


2 


: 


± 




'-. 


1 % HI 


9 : 


" : - 


-: 


m 


: - 


1% 2H 


--■- 


^: 


5 


- : ; 


- 


1% 'h\ 


- 


*': 


-- 


415 


- 


■2 2% 


334 


- 


-: 


'--. 


: '- 


'^A 


±* 


10 


6 


'- 


- 


2y 2 


-; 


-i 


10*6 


' : - 


' - 


- 


z% 


Hi 


&A 


10H 


'-: 


"rv 


': 


3 


4 


0% 


n# 


: 


?M 


": 


*Va 


-- 


' \ 


11^ 


'-- 


*A 


■ = 


$% 45: 


•?--. 


12 # 


-'-: 


8^ 


H 


3^ 


Hi 


Hf 


12*6 


7H 


• J ^ 


1 


4 


5^ 


7X 


13 


- 


10 


1 



GEAR TEETH. 375 

GEAR TEETH. 

Circular Pitch. 

The length of the pitch circle or pitch line from center of 
one tooth to the center of the next is the circular pitch of a gear, 
or a rack. 

Cast gear teeth, constructed on the circular pitch system, 
may be made of the following proportions : 

Thickness of tooth on pitch line = T 6 j pitch. 

Space between teeth on pitch line = T V pitch. 

Height of tooth outside of the pitch line = T 3 o pitch. 

Depth of space inside of pitch line = T % pitch. 

_,. , ,. . circular pitch X number of teeth. 
Pitch diameter of gear = - 



3.1416 

For cut gears, use the following formulas : 

Thickness of tooth on pitch line = 0.5 pitch. 
Space between teeth on pitch line = 0.5 pitch. 
Height of tooth outside pitch line = 0.3183 pitch. 
Depth of space inside of pitch line = 0.3683 pitch. 

To Calculate Diameter of Gear According to Circular Pitch. 

Rule. 

Multiply the circular pitch by the number of teeth in 
the gear and divide the product by 3.1416 ; the quotient is 
the diameter of the pitch circle ; add T 6 <y of the circular pitch to 
obtain the whole diameter of the gear. 

Example. 

Find whole diameter of a gear of 48 teeth and three-inch 
circular pitch. 

Solution : 

o X 48 
Pitch diameter = n * „ n = 45.84 inches. 
3.1416 

Double the addendum = 3 X 0.6 = 1.80 



The whole diameter is 47.64 inches. 

Table No. 47 is calculated for one-inch circular pitch ; to 
find the pitch diameter of a gear of any number of teeth given 
in the table, multiply the diameter given in the table by the 
circular pitch in the gear, and the product is the pitch diameter 
of the gear. In order to find the whole diameter, add twice the 
height of the tooth outside the pitch line, as calculated by the 
above formula. 



376 



GEAR TEETH. 



TABLE No. 47.— Giving Pitch Diameter of Gears of One 
Inch Circular Pitch. 



Teeth. 


Dia. 


Teeth. 


Dia. 


Teeth. 


Dia. 


Teeth. 


Dia. 


12 


3.82 


30 


11.46 


60 


19.10 


84 


26.74 


13 


4.14 


37 


11.78 


61 


19.42 


85 


27.06 


14 


4.46 


38 


12.10 


62 


19.74 


86 


27.38 


15 


4.78 


39 


12.42 


63 


20.06 


87 


27.70 


16 


5.09 


40 


12.73 


64 


20.37 


88 


28.01 


17 


5.41 


41 


13.05 


65 


20.69 


89 


28.33 


18 


5.73 


42 


13.37 


66 


21.01 


90 


28.65 


19 


6.05 


43 


13.69 , 


67 


21.33 


91 


28.97 


20 


6.37 


44 


U 


68 


21.65 


92 


29.29 


21 


6.69 


45 


14.32 ! 


69 


21.97 


93 


29.60 


22 


7 


46 


14.64 


70 


22.28 


94 


29.92 


23 


7.32 


47 


14.96 i 


71 


22.60 


95 


30.24 


24 


7.64 


48 


15.28 ; 


72 


22.92 


96 


30.56 


25 


7.96 


49 


15.60 


73 


23.24 


97 


30.88 


26 


8.28 


50 


15.92 ; 


74 


23.56 


98 


31.20 


27 


8.60 


51 


16.24 


75 


23.88 


99 


31.52 


28 


8.91 


52 


16.55 


76 


24.19 


100 


31.83 


29 


9.23 


53 


16.87 


77 


24.51 


101 


32.15 


30 


9.55 


54 


17.19 


78 


24.83 


102 


32.47 


31 


9.87 i 


55 


17.51 


79 


25.15 


103 


32.78 


32 


10.19 


56 


17.83 


80 


25.47 


104 


33.10 


33 


10.50 


57 


18.14 


81 


25.79 


105 


33.42 


34 


10.82 


58 


18.46 


82 


26.10 


106 


33.74 


35 


11.14 


59 


18.78 


83 


26.42 


107 


34.06 



To Calculate Diameter of Gears when Distance Between 
Centers and Ratio of Speed is Given. 

When calculating gears to connect two shafts of given dis- 
tance between centers and at a given ratio of speed, use the 
formula, 

_ 2 X S X n 



d = 



n + JV 

2X S XJV 

n + JV 



D = Diameter of large gear. 
d = Diameter of small gear. 
6" = Distance between centers in inches. 
JV= Number of revolutions of large gear per minute. 
n = Number of revolutions of small gear per minute. 
Note. — The small gear is always on the shaft having the 
greater speed. 



gear teeth. 377 

Example. 

What will be the diameter of the gears to connect two shafts 
when the distance between centers is 32 inches, and one shaft 
is to make 135 revolutions and the other 105 revolutions per 
minute ? 

Solution : 

2 X 32 X 135 „„-,,. 

L> = — —^z — ;— 77^ — = 36 inches diameter. 
13o -f- lOo 

, 2 X 32 X 105 nn . . 
a = — — — — t^tt^ — = 28 inches diameter. 
13d + lOo 

After the diameter of the gears is calculated, the pitch is 
decided upon according to the power the gears have to transmit. 

Frequently the pitch will have to be altered somewhat, and 
such gears sometimes have teeth of very odd pitch, in order to 
obtain the right number of teeth to give the required ratio of 
speed. The ratio between the number of teeth in the gears may 
always be seen from the ratio of speed between the two shafts. 
For instance, in the above example, the ratio of speed between 
the shafts is 13 ^io5, which, reduced to its lowest terms, is % ; 
therefore, the number of teeth in the two gears may be any 
multiple of 9 and 7, respectively. 

For instance, 8 X 9 = 72 teeth for the large gear, and 8 X 
7 — 56 teeth for the small gear ; or, 10 X 9 = 90 teeth for the 
large gear, and 10 X 7 = 70 teeth for the small gear, etc. 

The dimensions of teeth may be calculated according to 
rules given on page 375. 

Diametral Pitch. 

The dia7netral pitch of a gear is the number of teeth 
to each inch of its pitch diameter. In cut gearing it is always 
customary to calculate the gears according to diametral pitch. 
When gears are calculated according to circular pitch the corre- 
sponding circumference of the pitch circle is usually an even 
number, but the diameter will generally be a number having 
cumbersome fractions, and therefore the distance between the 
centers of the gears will be a number having fractions which 
may be very inconvenient to measure with common scales. 
This is because the circumference of a circle divided by 
3.1416 is equal to its diameter and the diameter multi- 
plied by 3.1416 is equal to the circumference. When 
gearing is calculated according to diametral pitch this trouble 
is entirely avoided, as this directly expresses the number of 
teeth on the circumference of the gear according to its pitch 
diameter. For instance, " six diametral pitch " means that 
there are six teeth on the circumference of the gear for each 
inch of pitch diameter. Thus, a gear of six diametral pitch and 
forty-eight teeth will be eight inches pitch diameter. A gear of 
" eight diametral pitch " means that the gear has eight teeth per 



. 



S GEAR TEETH. 



inch of pitch diameter, A gear of " ten diametral pitch n means 
that tie ^ear aas tea teeth per in :i i-frhtzi aiazieter. A rear 
of u twelve diametral pitch " means that the gear has twelve 
teeth per inch of pitch diameter, etc 

Thus, the pitch diameter and, consequently, the distance 
between the centers, will be a number which may be conven- 
iently measured, and the dimensions of tooth parts are also 
much more easily calculated by this system. 

Rules for Calculating Dimensions of Gears According to 
Diametral Pitch. 

The pitch diameter is obtained by dividing the number of 
teeth by the diametral pitch. 

Example. 

What is the pitch diameter of a gear of 43 teeth, 16 pitch ? 
Solution : 

48 divided by 16 = 3, therefore the pitch diameter is 3 
inches. 

The number of teeth is obtained by multiplying the pitch 
diameter by the diametral pitch. 

Example. 

\\~hat is the number of teeth in a gear of 5 inches pitch 
diameter and 12 pitch ? 
Solution : 
5 X 12 = 60, therefore the gear has 60 teeth. 

The whole diameter of a spur gear is obtained by adding 
2 to the number of teeth and dividing the sum by the diametral 

pitch. 

Example. 

What is the whole diameter of a gear blank for 68 teeth, 

10 pitch ? 

Solution : 

ft* -+- *> 

V. :.:.t aiarr.eter = — — — = 7 mcaes, 

The number of teeth is obtained by multiplying the whole 
diameter of the gear by the diametral pitch and subtracting 2 
from the product. 

Example. 

The whole diameter of a gear blank is 8 inches : it is to 
be cut 10 diametral pitch. Find the number of teeth. 
Solution: 
Number of teeth = (8 X 10) — 2 = 78. 

The diametral pitch is obtained by adding 2 to the number 
: : teeth and dividing by the whole diameter. 



gear teeth. 379 

Example. 

A gear has 64 teeth and the whole diameter is 16>£ inches. 
What is the diametral pitch ? 
Solution : 

~. . . . 64 + 2 

Diametral pitch = - -—y- = 4. 

16K 
Thus, the gear is 4 diametral pitch. 

Note. — The term diameter of a gear usually means diame- 
ter of pitch circle. 

The distance between the centers of two spur gears is ob- 
tained by dividing half the sum of their teeth by the diametral 
pitch. 

Example. 

What is the distance between centers of two gears of 48 
and 64 teeth and 8 diametral pitch ? 
Solution : 

Distance = = 7 inches. 

2X8 

The circular pitch is obtained by dividing the constant 
3.1416 by the diametral pitch. 
Example. 

What is the circular pitch of a gear of eight diametral 
pitch ? 

Solution : 

Circular pitch = — '— — = 0.393 inch. 

o 

The thickness of the tooth on the pitch line is obtained by 
dividing the constant 1.5708 by the diametral pitch. 
Example. 

What is the thickness of the tooth on the pitch line of a 
gear of 6 diametral pitch ? 
Solution : 

1 K>7AG 

Thickness of tooth = - L - j - — = 0.262 inches. 

o 

The working depth of the tooth is obtained by dividing 2 
by the diametral pitch. The clearance at the bottom of the 
teeth is T ! o of the thickness of the tooth on the pitch line. The 
whole depth to cut the gear is obtained by dividing the constant 
2.157 by the diametral pitch. 

Example. 

Find the depth to cut a gear of 8 diametral pitch. 
Solution : 

Depth = ^4°^ = O- 2 ? ^ch. 



3 8o 



GEAR TEETH. 



The whole depth is nearly equal to 0.0866 times the circular 
pitch. The use of the following tables will facilitate calcula- 
tions regarding dimensions of teeth in diametral pitch. 



TABLE No. 48. — Comparing Circular and Diametral Pitch. 



Diametral Pitch. 


Circular Pitch. 


Circular Pitch. 


Diametral Pitch. 


2 


1.571 inch. 


\y 2 inch. 


2.094 


2^ 


1.257 " 


1 T 7 6 


u 


2.185 


3 


1.047 " 


in 


a 


2.285 


sy 2 


0.898 " 


lj 5 6 


(C 


2.394 


4 


0.785 " 


IX 


it 


2.513 


5 


0.028 " 


l T 3 e 


it 


2.646 


6 


0.524 " 


1/s 


a 


2.793 


7 


0.449 " 


ItV 


a 


2.957 


8 


0.393 " 


1 


a 


3.142 


9 


0.349 " 


15 
1 6 


a 


3.351 


10 


0.314 " 


n 


a 


3.590 


11 


0.286 " 


13 
1 6 


a 


3.867 


12 


0.262 " 


u 


a 


4.189 


14 


0.224 " 


1 1 

1 6 


a 


4.570 


16 


0.196 " 


5 A 


a 


5.027 


18 


0.175 " 


9 

1 6 


a 


5.585 


20 


0.157 " 


% 


a 


6.283 


22 


0.143 " 


7 

1 6 


a 


7.181 


24 


0.131 " 


H 


a 


8.378 


26 


0.121 " 


5 

1 6 


a 


10.053 


28 


0.112 " 


X 


it 


12.566 


30 


0.105 " 


3 
1 6 


a 


16.755 


32 


0.098 " 


X A 


it 


25.133 


TABLE No. < 


19. — Giving Dimensions of Teeth Calculated 


j 


\ccording to Diametral Pitch. 


Diametral 
Pitch. 


Depth 
Cut in 


to be 
Gear. 


Thickness of 

Tooth on 
Pitch Line. 


Diametral 
Pitch. 


Depth to be 
Cut in Gear. 


Thickness of 

Tooth on 

Pitch Line. 


2 


1.078 


in. 


0.785 in. 


12 


0.180 in. 


0.131 in. 


2/ 2 


0.863 




0.628 


14 


0.154 


0.112 


3 


0.719 




0.523 


16 


0.135 


0.098 


sy 2 


0.616 




0.448 


18 


0.120 


0.087 


4 


0.539 




0.393 


20 


0.108 


0.079 


5 


0.431 




0.314 


22 


0.098 


0.071 


6 


0.359 




0.262 


24 


0.090 


0.065 


7 


0.307 




0.224 


26 


0.083 


0.060 


8 


0.270 




0.196 


28 


0.077 


0.056 


9 


0.240 




0.175 


30 


0.072 


0.052 


10 


0.216 




0.157 


32 


0.067 


0.049 


11 


0.196 




0.143 









GEAR TEETH. 38 1 

To Calculate the Number of Teeth when Distance Be= 
tween Centers and Ratio of Speed is Given. 

Select for a trial calculation, the diametral pitch which 
seems most suitable for the work. 

Calculate the sum of the number of teeth in both gears 
corresponding to this pitch by multiplying twice the distance 
between their centers by the diametral pitch selected. 

The number of teeth in each gear is obtained by the follow- 
ing formula : 

NX A 



T — 



t = 



n+AT 

n X A 

JV+ n 



T = Number of teeth in large gear. 

/ = Number of teeth in small gear. 
AT=- Number of revolutions of small gear. 
n ==■ Number of revolutions of large gear. 
A = Number of teeth in both gears. 

Example. 

The center distance between two shafts is 15 inches. The 
small gear should make 126 and the large gear, 90 revolutions 
per minute. Calculate the number of teeth in each gear, if 8 
diametral pitch is wanted. 

Solution: 

The number of teeth in both gears is 2 X 15 X 8 = 240. 

T= 126X240 =140 teeth. 
126 + 90 

t = 90 X 240 = 100 teeth. 
126 + 90 

Frequently it is impossible to get gears of the desired pitch 
to fit within the given center distance and to give the exact 
ratio of speed. Some modifications must then be made ; either 
the exact ratio of speed must be sacrificed, the pitch must be 
changed, or the distance between centers must be altered. 

Note. — The ratio of the number of teeth in the gears can be 
seen from the ratio of the speed. For instance, in the above ex- 
ample the ratio of speed is 9 %26, which, reduced to its lowest 
terms, is % ; therefore, the number of teeth in the two gears 
may, with regard to speed ratio, be any multiple of 5 and 7, 
respectively, but in order to fit the given center distance and 
also to be 8 pitch, they must be 100 and 140, which is 20 X 5 
= 100 and 20 X 7 = 140. 



GEAR TEETH. 




The shape of gear teeth is usually either Involute or Cycloid 
( also frequently called Epicycloid ). The shape of a cycloid 
tooth for a rack is four equal cycloid curves, which may be con- 
! tructed, so to speak, by letting the generating circle a, ( see 
Fig. 1 ) roll along on the pitch line of the rack, both above and 
below. 

Cycloid gears 
have the curve out- 
side the pitch circle 
formed by an Epi- 
cycloid (see Fig. 26, 
page 191) and the 
curve inside the 
pitch circle by a 
Hypocycloid. 

Trie curves al- 
ways meet on the 
pitch line in both 
gears and racks. 

The theoretical 
requirements for 
correct form of Epi- 
cycloid gear teeth 
are that the face of 
the teeth of one gear and the flank of the teeth of the other gear 
must be produced by generating circles of the same diameter. 

The diameter of the generating circle is limited by the size 
of the smallest gear or pinion in the series of gears which are con- 
structed to run together, because if the generating circle is as 
large in diameter as half the pitch diameter of the gear, the 
hypocycloid will be a straight line ; thus, the flank of the tooth 
will be a straight radial line. If the generating circle is 
larger than half the pitch diameter of the gear, the result will be 
a weak and poor tooth with under-cut flank. 

When the same size of generating circle is used for gears of 
different diameters but of the same pitch, all such gears will work 
correctly together, and for this reason it is possible to construct 
interchangeable gears having cycloid teeth. If the diameter of 
the generating circle is equal to half the diameter of the 
smallest gear in the set, this gear will have teeth with radial 
flanks but all the other gears and the rack will have double- 
curved teeth. Fig. 1 shows a rack drawn to >£-inch circular 
pitch : the generating circle is 0.98 inch diameter, which is equal 
to half of the pitch diameter of a gear of 12 teeth and % inch 
circular pitch, 

All gears of the same pitch having 12 teeth or more, con- 
structed by the same generating circle in the same manner as 
the rack, will match and be interchangeable with the rack, and 
will also match and be Interchangeable with each other. 



GEAR TEETH. 383 

When internal teeth are constructed by the above method, the 
difference between the number of teeth in the internal gear and 
its external pinion must never be less than 12 ; practically, it is 
better to limit the difference to 15 or 20 teeth. 

As interchangeability is seldom required for internal gear- 
ing, such gears and their mates are generally constructed together 
and the designer chooses a generating circle of suitable size to 
give the shape of tooth he considers best, and he may also vary 
the size of the driving or the driven gear so as to reduce con- 
tacts when the teeth are approaching each other, etc., according 
to his own judgment and experience. 

The difference in pitch diameter of the internal gear and its 
pinion should never be less than the sum of the diameters of the 
generating circles, and the diameter of the generating circle of 
the flanks for the pinion should never be larger than half the 
pitch diameter, but it should, preferably, be smaller. 

As a rule, fillets at the bottom of the teeth are not used in 
internal gears, but if used they should be very small. 

In order that gears constructed with cycloid teeth should 
run smoothly, it is very important to have the distance between 
centers correct, so that the pitch lines will exactly meet each 
other. For this reason, there are many kinds of machinery 
where cycloid gears should not be used : for instance, for change 
gears on lathes, involute teeth as far more suitable. 

When making patterns, the shape of one tooth is usually 
carefully drawn on a thin piece of sheet metal, either brass or 
iron ; this is then filed out and used as a templet in tracing 
the other teeth on the pattern. Sometimes a fly-cutter is made 
according to this constructed tooth, and all the teeth in the pat- 
tern are cut on an index machine or a gear cutting machine ; but 
if such a machine is not available, the next best way is to set out 
the pitch line of the gear on this templet and also the center line 
of the tooth, radially towards the center, then draw the pitch line 
on the pattern, space off each tooth carefully with a pair of 
dividers and draw the center line on each tooth prolonged across 
the rim radially in the direction of the center of the gear, then 
lay the templet carefully on each of these spacings, making 
the pitch line and the center line of tooth on the templet to 
exactly match the pitch line and center line of the tooth 
drawn on the pattern, then trace around the templet and get the 
shape of one tooth ; then move the templet to the next spacing 
and trace the next tooth, and so on for all the teeth on the gear. 

For small patterns it is convenient to fasten the templet to 
a strip of metal long enough to reach from the teeth to the cen- 
ter of the gear wheel, placing a point in the center of the gear, 
drilling a hole in the strip and letting it swing around this point, 
then after all the teeth are spaced off on the pattern the tem- 
plet is swung from one tooth to the other and all the teeth are 
traced by the templet. This method has the advantage that 



384 GEAR TEETH. 

it will mark all the teeth exactly alike, because .the templet 
being fastened to this strip, can not easily get out of position. 

The distance from the pitch line of the templet to the cen- 
ter hole in the strip must be laid off according to the shrinkage 
rule, and is, of course, in numerical value equal to the pitch radius 
of the gear, which should always be calculated and given on the 
drawing. When gear patterns are less than six inches in diame- 
ter it is preferable not to allow anything for shrinkage, as the 
moulder will usually rap the pattern about as much as the cast- 
ing will shrink in cooling. 

When a pair of cycloid gears are constructed without con- 
sidering interchangeability with other gears of the same pitch, 
it is customary to choose a generating circle having a diameter 
equal to three-fourths of the radius of the pitch circle of the 
small gear, providing this gear has 24 or more teeth. A large 
generating circle probably reduces the friction in a small 
measure but gives teeth of less strength. The largest gen- 
erating circle used ought never to exceed the radius of the 
pitch circle of the small gear. Decreasing the generating 
circle will probably increase friction somewhat in the gears, but 
it gives teeth of greater strength. The smallest generating circle 
used in practice is equal to half the diameter of the pitch circle 
of a gear having 12 teeth of the same pitch as the gear to be con- 
structed. Many eminent mechanics consider it preferable 
never to use a generating circle smaller than half of the pitch 
diameter of a gear of 15 teeth. 

Cycloid gears are mostly used in large cast gears of one- 
inch circular pitch or more. 

Sometimes the driving gear is made of slightly larger di- 
ameter, and the teeth spaced at a correspondingly greater pitch 
than the theoretically calculated size. This is done in order 
that the teeth shall not rub on each other on the approaching 
side, but only touch as they are passing the center line and 
commence to slide away from each other. This will make the 
gears less noisy, but probably gears made in this manner will 
wear faster, as there are fewer points of contact, although this 
may be offset by the fact that the friction between the teeth 
when the}- are meeting and pushing onto each other is more 
injurious than the friction produced when they are sliding away 
from each other. 

The same idea is sometimes employed when constructing 
bevel gears, in order to make them run quietly. 

This mode of sizing gears is not. as a rule, used in modern 
gear construction, but it is a point well worth remembering, 
because if either of two gears is over or under size, the gear of 
over-size should always be used as the driver, and the gear of 
under-size should always be the driven; never vice versa. This 
will apply as well to involute as to epicycloid gears. 



GEAR TEETH. 



3S5 



Involute Teeth. 

Suppose a strap is fastened at a and b on the two round discs 
in Fig. 2. If the disc b is turned in the 
direction of the arrow, the strap will move 
in a straight line from c, toward d. This 
motion will cause the disc a to rotate with 
exactly the same surface speed as the disc 
b, but in the opposite direction. 

Suppose, further, that to the under side 
of the disc a (see Fig. 3) is fastened a piece 
of sheet brass p, or other suitable material 
of somewhat larger diameter than disc a, 
and that a scratch awl is fastened in the 
strap at the point m ; then by turning the 
disc b in the direction of h to b, and 
the strap moving with it, being kept tight 
by the resistance of disc a, the scratch awl 
will trace on the brass plate the curve from 
m to h, but if the discs are moving in the 
opposite direction, the scratch awl will 
trace the curve from m to K. Take an- 
other brass plate and do the same thing with the other disc, and 
a similar curve will be produced. In these two brass plates the 
stock may be filed away carefully, following the curves as shown 
in Fig. 4. The discs are laid to match each other and free to 




Fig. 2. 





Fig. 4. 



Fig. 3. 



swing on their centers ; turning the disc a in the direction of the 
arrow, it will give motion to b, and both discs will move with the 
same speed in exactly the same manner as if they were connected 
by the strap as shown in Fig. 2. 



3 86 



GEAR TEETH. 



The curve on these two discs represents the form of a gear 
tooth in the involute system.* 

The line h g, Fig. 4, is called the line of pressure or the line 
of action. The circles, P and P, are the pitch circles. The line 
B P, shows the direction of motion of the teeth at the moment 
they are passing the center line, c c. 



F..G: v 5. 



Approximate Construction of Involute Teeth. 

It will be noticed that the line of pressure, h g forms an 
angle with the line B P. This angle is usually taken as 14^ 
degrees. This makes the diameter of the base circle, £, (see 
Fig. 5) equal to 0.968 times the diameter of the pitch circle. 
The base circle g-g, in Fig. 5, corresponds to the disc in Fig. 3, 
and the line of pressure in Figs. 5 and 6 corresponds to the strap 
in Fig. 2. The line of pressure, h g. Fig. 5, is 75 l /z degrees to 
the center line, f c. 

A perpendicu- 
lar is erected 
from the line h 
g, through the 
center, c. Using 
the point of in- 
tersection at / as 
center, the tooth 
is drawn simply 
by a circular arc. 
This will, in prac- 
tical work for 
small gears hav- 
ing more than 
twenty teeth, cor- 
respond nearly 
enough to the 
true involute, which was illustrated by means of the strap, disc 
and scratch awl, as explained in Figs. 2, 3 and 4. 

When the gear has less than twenty teeth, and is constructed 
by circular arcs, as shown in Fig. 5, the top of the tooth will be 
too thin ; but the top of the tooth will be too thick to clear in 
the rack, if the true involute curve is used. 

When the teeth are of true involute curve, a smaller gear 
than twenty-five teeth will not run freely in a rack having straight 
teeth slanting 14^ degrees. (See Figs. 6 and 7). Therefore, 




*The way to actually draw this curve on paper by means of drawing instru- 
ments is explained on page 192. This way explained here, using the disc on the 
strap, is merely for illustrating and explaining principles, and serves well for that 
purpose, but would be inconvenient to use in actual construction of gear teeth. In 
actual work one tooth is carefully constructed, and templets and cutters are made 
and used, as was explained for Cycloid Gears, pa^e 383. 



GEAR TEETH. 



387 



when a gear has less than twenty-five teeth it is necessary to round 
the teeth somewhat outside the pitch circle. By making either a 
drawing or a templet, it is very easy to see how much to round 

J\ 

>/! 



&-^ 




Involute Teeth (Cast.) 



the teeth to make them clear in the rack. In interchangeable 
sets of cut involute gears it is customary to cut the rack with a 
cutter shaped for a gear of 135 teeth. This will make the teeth 
in the rack slightly curved instead of straight, as shown in Fig. 6, 
and this will also make it possible to construct the small gears in 
an interchangeable set nearer to a true involute, and still have 
them run freely in the rack. 



388 



GEAR TEETH. 



When gear teeth are constructed as shown in Figs. 6 and 7, 
the line g h, is 15% degrees to the line c f, and the line c z\ is 
14>^ degrees to the line c f. (See Fig. 5). 




s*jk>. 



Fig. 7. Involute Teeth (Cut). 

The line h g, will always be tangent to the base circle 
which is concentric to the pitch circle. The diameter of the 
base circle is always 0.9G8 times the diameter of the pitch circle. 
The circle forming the shape of the tooth must always have its 
center on the circumference of the base circle, and its diameter 
will be one-fourth of the pitch diameter of the gear. As shown 
in Fig. 2, the same circle gives the form of tooth for coarser or 
finer pitch. When gears are drawn by this method the pitch 
circle is divided into as many teeth and spaces as there are to 
be teeth in the gear ; then the form of the tooth is simply struck 
by the dividers, always using the periphery of the base circle as 
center, and always taking the distance in the dividers equal to 
one-fourth of the radius of the pitch circle. 

The diameter of the base circle is 0.968 times the diameter of 
the pitch circle, because cosine of 14^ degrees is 0.96815. The 
diameter of the circle forming the shape of the tooth is 0.25 times 
the diameter of the pitch circle of the gear, because sine of 14>£ 
degrees is 0.25038. If the line of pressure is laid at any other angle 



GEAR TEETH. 389 

than 14% degrees, all these other proportions will also change. 
Fig. 6 shows a pattern for gears and rack constructed with 
necessary clearance as used for cast gears. All tooth parts are 
of the same dimensions as used for cycloid gears as given on 
page 375. Fig. T shows a cut gear and rack constructed in the 
same manner. The advantages of the involute system of gears 
are in the strength of teeth, and also that the gears will trans- 
mit uniform motion and run satisfactorily, even if the distance 
between centers should be slightly incorrect. 

Width of Gear Wheels. 

Gears with cast teeth are usually made narrower than gears 
with cut teeth. In spur gears with cast teeth it is customary to 
make the width of the gear four to five times the thickness of 
the teeth, or twice the circular pitch. 

Width of Gears With Cut Teeth. 

The following rule is recommended by Brown & Sharpe 
Mfg. Co. in their " Practical Treatise on Gearing": 

Divide eight by the diametral pitch, and add one-fourth inch 
to the quotient; the sum will be the width of face for the pitch 
required. 

Example. 

What width of face is required for a gear of four pitch ? 

Solution : 

Face = ! + ? = ~ l A inches. 
For change gears on lathes where it is desirable not to have 
faces very wide, the following rule may be used : 

Divide four by the diametral pitch and add one-half inch. 
By the latter rule a four-pitch change gear would have but 
a 1%-inch face. 



BEVEL GEARS. 

Fig. 8 is a diagram showing how to size bevel gear blanks. 

First, lay off the pitch diameters of the two gears, which 
may be calculated according to diametral pitch or to circular 
pitch ; second, draw the pitch line of teeth ; third, lay off on the 
back of the gear the line a &, square to the ' ; pitch line of teeth ;"' 
fourth, on the line a b, lay off the dimensions of the teeth exactly 
in the same manner as if it was for a spur gear. 

If the gear is calculated according to circular pitch, find 
dimensions of teeth by formulas on page 875, but if the gear is 
calculated according to diametral pitch, find dimensions of teeth 
in Table No. 49. 



390 



BEVEL GEARS. 



Make the drawing carefully to scale ( full size preferable 
whenever possible ), and measure the outside diameter as shown 
in the diagram. 




FIG. 8. 



To Calculate Size of Bevel Gear for a Given 
Ratio of Speed. 

Ascertain the ratio of speed in its lowest terms. Multiply 
each term separately by the same number, and the products 
give the number of teeth in each gear. 

Example. 

Two shafts are to be connected by bevel gears, one shaft to 
make 80 revolutions and the other 170 revolutions per minute. 
Find the number of teeth in the gears. 



BEVEL GEARS. 



391 



Solution : 



Ratio = 80 /i70 = s /i7. 



For instance, multiplying by 6, the large gear on the shaft 
making 80 revolutions will have 17X6 = 102 teeth. The small 
gear on the shaft making 170 revolutions will have 8 X 6 = 48 
teeth. 

Assuming that on account of room it is necessary to use 
smaller gears, a smaller multiplier may be used, but if it is desir- 
able to have larger gears, use a larger multiplier. 

Decide on the pitch of the gears according to the work they 
are required to do. Make a scale drawing and get the 
dimensions as explained on page 390. 

Dimensions of Tooth Parts in Bevel Gears. 

Fig. 9 shows a sectional drawing of a pair of bevel gears of 
sixteen diametral pitch, 18 teeth in the small gear and 30 teeth 
in the large gear. The pitch diameter of the small gear is ||- = 
1% inches. The pitch diameter of the large gear is ff = 1% 
inches. 



FIG. 9 




The addendum of the teeth on the back at a is y 1 ^ inch 3 
the same as for a spur gear of 16 diametral pitch. The thickness 



392 BEVEL GEARS. 

and the total depth to cut the gear at a are 0.098 inch and 0.135 
inch, respective^". 

These dimensions are found in Table No. 49, as if it was 
a spur gear of 1$ diametral pitch. All the dimensions of the 
tooth decrease gradually toward b, as the whole tooth is sup- 
posed to vanish in a point in the center at c. The dimensions 
of the teeth at b may be calculated and are always in tke same 
proportion to the dimensions at a as the distance c b is to the 
;. s :ance c a ; th . s, if t b e length of the tooth from a to b is made 
one-third of the length of the distance c a. the distance b c is 
two-thirds of the iistance a c. and. consequently, all the dimen- 
sions of the tooth at b are two-thirds of the dimensions at a. 
Instead of calculating the size of die teeth at b, the dimensions 
may be obtained by careful drawing. The depth of the tooth at 
the smallest end is then measured darecdy at b, but die thickness 
is measured at /; the distance / h is laid off equal to b d. 

The length of the tooth from a to b is to a certain extent 
arbitrary, but a good rule is seven inches divided by the diame- 
tral pitch, but never longer than one-third of the distance from 
a to :. 

Example. 

What is the proper length for the teeth of a bevel gear : : B 
diametral pitch ? 

Solution: 

Seven inches divided by 8 = % inch, if the gears ire of such 

diameters that this will not make the length of the teeth more 

than one-third of the distance from a to c. 



Form of Tooth in Bevel Gears. 

Extend the line a (see Fig. 9). until it intersects the axial 
center line of the gear, as at I \\ use i is hie center, and the shape 
of tooth at a for the large gear is constructed as if it was a spur 
gear having a pitch radius as large as a k. 

The shape of the tooth at b is constructed in the same way. 
by extending the line b. (which always — the same as line a. — is 
square to the pitch line of the tooth) until it intersects the 
axial center line of the gear, as at .:'. Using d is center, the 
shape of the tooth is constructed as if i: was i spur gear having 
a pitch radios e :ual to db. The shape of the teeth of the small 
gear is obtained in the same way. which is shown by the drawing. 

The form of tooth is shown to be approximately involute, 
constructed as explained for spur gears, | ire 886. 

Measuring the back cone radios, a ':. :i the large geir. i: :s 
found to be f| inch, and the diameter will be f§ inch : hi is. the 
shape of the tooth at a for the large gear will be the same as the 
shape of the tooth in a spur gear of 58 teeth, sixteen liametral 
pitch. 



BEVEL GEARS. 393 

Measuring the back cone radius of the small gear, it is 
found to be f h inch, and the diameter will be f | inch ; conse- 
quently the shape of tooth at a for the small gear is the same as 
the shape of tooth in a spur gear of 21 teeth, sixteen diametral 
pitch. 

Therefore, if this pair of gears is to be cut by a rotary 
cutter having a fixed curve, a different cutter is required for 
each gear. 

When, in a pair of bevel gears, both gears are of the same 
size and have the same number of teeth, and their axial center 
lines are at right angles, they are called miter gears, and one 
cutter, of course, will answer for both gears. One cutter will 
also answer in practice when the difference of the back cone 
radius of a pair of gears is so small that it comes within the 
limit of one cutter as used for spur gears of the same size. Bevel 
gears may also be made with cycloid form of teeth, but when- 
ever cut by rotary cutters, as usually employed in producing 
small bevel gears of diametral pitch, the involute form of tooth 
should always be used. 

Cutting Bevel Gears. 

When bevel gear teeth are correctly formed, the tooth 
curve will constantly change, from one end of the tooth to the 
other. Therefore, bevel gears of theoretically correct form 
cannot be produced by a cutter of fixed curve ; but, practically, 
very satisfactory results are obtained in cutting bevel gears of 
small and medium size in this way. 

When a regular gear-cutting machine is not at hand, the 
Universal milling machine is a very convenient tool for cutting 
bevel gears of moderate size, and is used in the following way : 

First, see that the gear blank is turned to correct size and 
angle, and adjust the machine to the angle corresponding to the 
bottom of the teeth in the gear. The correct index is set ac- 
cording to the number of teeth in the gear. Adjust the cutter 
to come right to the centre of the gear, cut the correct depth as 
marked on the gear at a (see Fig. 9), according to Table No. 49, 
and when the machine is adjusted to the correct angle, and the 
correct depth is cut at #, the correct depth at b will, as a matter 
of fact, be obtained. 

Second, when a few teeth are cut in the gear (two or three) 
bring, by means of the index, the first tooth back to the cutter. 
By means ©f the index, rotate the gear, moving the tooth toward 
the cutter ; but, by the slide, move the gear sidewise away from 
the cutter, until the cutter coincides with the space-at b \ then cut 
through from a to b. This operation will widen one side of the 
tooth space at a. 

Note the position of the machine, and, by the use of the 
jndex and slide, return the cutter to its central position and in- 



394 BEVEL GEARS. 

dex into the next space, and rotate the other side of the tooth 
toward the cutter as much as the first side ; but. by the slide, 
the gear is moved side wise away from the cutter until the 
cutter coincides with the space at bj then cut through on this side 
from a to b. Thus, by repeated cutting on each side alternate- 
ly, one tooth is backed off equally on both sides and measured 
by a gage, until the correct thickness on the pitch-line at a, 
according to Table No. 49, is obtained. 

Be very careful to have the machine set over the same 
amount on each side of the tooth, or else the tooth will be 
askew. 

Third, when one tooth, thus by trial, is correctly cut, 
note the position of the machine and cut all the teeth through 
on one side, then set over to the other side in exactly the 
same position as was found to be right for the first tooth ; cut 
through again and the gear is finished. Thus, when the correct 
position of the machine is obtained, any number of gears of 
the same size and same pitch may be cut, by simply letting the 
cutter go through twice. 

Note. — As already stated, bevel gear cutting in this way is 
only a compromise at the best, but by careful manipulation 
and good judgment an experienced man is able to do a very 
creditable job. A cutter is usually selected of the same curve 
as is correct for a spur gear corresponding to the back cone 
radius of the gear. Thus, it may be thought that the shape 
of the tooth should be the shape of the cutter, but by investi- 
gation it will be found that, on account of the " backing off," 
the teeth will be of a little more rounding shape at the large end 
than corresponds to the cutter ; therefore, when the gear has few 
teeth. — less than 25. — it is usually preferable to make the shape 
of the cutter to correspond to a gear a little larger than would 
be called for by the back cone radius of the bevel gear to be 
cut; but when the gear has more than 25 teeth, a cutter of shape 
corresponding to the back cone radius of the gear will give 
good results. For instance, in the pair of bevel gears shown in 
Fig. 9 the back cone radius of the large gear calls for a 
cutter corresponding in shape to a cutter for a spur gear of 59 
teeth, 16 diametral pitch ; and this shape of cutter will, after 
the teeth are backed off, make the teeth a trifle too round at 
the large end, and a trifle too straight on the small end, but if 
the teeth are not too long the job will be very satisfactory. 

The back cone radius of the small gear calls for a cutter 
corresponding in shape to a cutter for a spur gear of 21 teeth, 
16 diametral pitch, but when the teeth are backed off they 
will be a little too rounding on the large end ; therefore a better 
result is obtained by selecting a cutter having a shape corre- 
sponding to a little larger spur gear ; for instance, a gear of 
24 teeth, Such a cutter will give the teeth a better shape on 



BEVEL GEARS. 



395 



the large end, although it may be necessary to round the teeth 
a little, outside the pitch line on the small end, by filing. 

Of course, a spur gear cutter cannot be used for cut- 
ting bevel gears, because, although it may have the correct 
curve, it would be too thick. The thickness of a bevel gear 
cutter must be at least 0.005 inch thinner than the space be- 
tween the teeth at their small end. 

Large bevel gears are made on theoretically correct prin- 
ciples by planing on specially constructed machines. 



WORMS AND WORM GEARS. 

Fig. 10 shows a worm and worm gear. 



TTt 




/=. Pitch diameter of gear. 

g — Smallest outside diameter. 

h — Largest outside diameter. 

a = Outside diameter of worm. 

b = Pitch diameter of worm. 

c = Diameter of worm at bottom of thread. 

The pitch and diameter of worm screws are usually of such 
proportions that for single-thread the angle of the teeth on the 
gear is from two to three degrees. This angle is most conven- 
iently obtained by drawing a diagram as shown in Fig. 10, 



39$ 



WORMS AND WORM GEARS. 



Draw a line ////, equal to 3} times the length of line b\ this line 
will be equal to the length of the circumference of the pitch 
diameter of the screw. Erect the perpendicular, m o, equal to 
the pitch of the screw. Connect the points / and o by the line 
I o, and the angle s is the angle of the teeth on the worm gear. 

Note. — When the pitch diameter of the worm screw is 
seven times the circular pitch of the worm gear and the worm 
has single thread, the angle of the thread on the gear is very 
nearly 2)4 degrees. 

Caution. — When cutting a worm gear, be careful and not 
lay the angle of the teeth in the wrong direction. 

The diameter of worm gears is usually calculated according 
to circular pitch, for convenience in cutting the worm with the 
same gears as used for ordinary screw cutting in a lathe. 

When a worm gear has comparatively few teeth, the flank 
of the tooth will be undercut by the hob; to prevent this in a 
measure, it is customary tojiave the blank somewhat over size, 
so that from five-eighths to three-fourths of the depth of the tooth 
may be outside the pitch line. 

The form of teeth is usually involute, and the thread 
on a worm screw is constructed of the same shape as the teeth 
in a rack. Fig. 11 shows the shape of tooth and the tabli 
gives the dimensions of finishing tool for the most common 
pitches. 

The surface speed of a worm screw ought not to exceed 800 
feet per minute. 

Table No. 50 is calculated by the following formulas. (See 
Fig. 11.) 




P- 


- Circular pitch. 


A r 


1 


l\ — 


P 


7\,T 


8.1416 


LVl 


P 


t — 


-T = s 


a = 


P X 0.3183 


d = 


■ P X 0.3683 


D = 


a + d 


S = 


P X 0.5 


b = 


P X 0.31 


C = 


P X 0.335 


k = 


D+J Y 


k = 


P x 0.1 



WORMS AND WORM GEARS. 



397 



TABLE No. 50.— Giving Proportions of Parts for Worms 
and Worm Gears, Calculated According to Circular Pitch. 

(See Fig. 11.) 



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worms axd worm gears. 



TABLE No. 51. — Showing How to Gear Lathes when Cut= 
ting Worms of the Pitches Given in Table No. 50. 



z 

— 

— 09 


a 

z _. 






LEADING SCREW OF LATHE. 


< 


4 


3 II 4 


5 6 10 


- z 




Threads 


Threads ! Threads 


Thre? Is 


Threads Threads 


- Z 
< z 

5 


5 - 

- - 

s 


per 


inch. 1 


per 1 


rich. per inch, 

1 


per inch. 


per inch, per inch. 


u 


i --. 




. \ . 


5 - 


_ - 


> - 

5 - 


s 








~ C 


V V 


~ V 


y u ~z « 


u u 


_ ■, 


■~ ■_ 


J u 


- • — i. I 


U 


z 


— zc 


- Z£ 

-J 


5 ai 


: - : = M 




E M 


-- -u 


3 zi 


5« 3M 








7 


X 


7. 


en 7. 


S. 


7. 


tffl 


7. 


x r. 


x 


2 


X 


160 


40 






1% 


% 


140 


40 




















iX 


X 


120 


40 






















IX 


% 


100 


40 


120 


32 


















1 


1 


80 


40; 


96 


32 


















X iX 


60 


40 


72 


32 120 


40 














% 1 3 5 


50 


40 


, 60 


32 100 


40 














X 2 


40 


40 


48 


32 


S( 


40 


100 


40 










% ?X 


40 


50 


48 


40 


64 


40 


80 


40 










x 3 


40 


60 


48 


48 


64 


4- 


60 


36 


48 


24 






-7 33< 


4'.' 


TO 


4S 


56 


64 


56 


60 


-- 


48 


28 






X 


4 ' 


40 


80 


24 


32 


40 


40 


50 


40 


48 


32 






% 


^x 


40 


90 


24 


36 


32 


36 


40 


3t3 


48 


36 






It 

O 


5 


20 


50 


24 


40 


32 


40 


40 


40 


48 


40 60 


30 


% 6 


20 


60 


24 


48 


32 


4^ 


40 


48 


48 


48 60 


36 


X 8 

A 10 


20 


80 


24 


64 


24 


48 


30 48 


48 


64 60 


48 


20 


100 


24 


80 24 


60 


24 48 


24 


40 50 


."■. 



Reduction of Speed by Worm Gearing. 

In a single-threaded worm, screw one revolution of the worm 
moves the gear one tooth : in a double-threaded worm screw one 
revolution of the worm moves the gear two teeth, and in a triple- 
threaded worm screw one revolution of the -worm moves the 
gear three teeth. A great deal of work is lost by friction by 
using worm gearing, frequently from 50 to T5 per cent., some of 
which could be saved by using a ball bearing to take the end 
thrust of the worm. The efficiency is also increased by using a 
worm of double, triple or quadruple thread, because this in- 
creases the angle of the teeth in the wheel and the efficiency of 
the mechanism is increased by increasing the angle until it 
reaches 20 c to 25°, when it rapidly falls oft again. 

Calculating the Size of Worm Gears. 

Example. 

Find dimensions of a worm gear having 6S teeth, % inch 
pitch, cut teeth. Make the pitch diameter of the single-thread 
worm six times the pitch of the worm. Use Table No. 50. 



WORMS AND WORM GEARS. 399 

Solution : 

In Table No. 47 the pitch diameter of a gear of 68 teeth of 
one-inch pitch is given as 21.65 inches; thus, the pitch diameter 
for a gear of 6S teeth of %-inch circular pitch will be 21.65 X 
0.75 = 16.738 inches. 

In column a, of Table No. 50, the addendum for %-inch 
circular pitch is given as 0.2387 inch ; fhis is multiplied by 2, 
because it is to be added on both sides of the gear. 

Thus, the smallest outside diameter of the gear is 16.738 -b 
0.2387 X 2 = 17.215 inches ; or, practically, 17g% inches. If the 
gear is to be made hollow to correspond to the curve at bottom 
of thread of the worm, make a scale drawing as shown in Fig. 
10, and make line g, 17g 7 2 inches ; from this drawing the largest 
outside diameter may be obtained by measurement. 

The diameter of the worm on the pitch line was to be six 
times the pitch = 6X^" = 4t]/ 2 inches. 

The addendum for the thread on the worm can be obtained 
from Table No. 50, column a, and is 0.2387. The outside 
diameter of the worm will be 4.5 + 2 X 0.2387 = 4.977 inches, or, 
practically, 4-f^ inches. 

The cutter to be used in roughing out the gear should have 
a curve of involute form corresponding to a spur gear cutter for 
68 teeth, and its thickness ought to be at least 0.005 inch less 
than the width of space as given in column 6^ of Table No. 50. 
Therefore the thickness on the pitch line of the roughing cutter 
will be 0.37 inch. 

The angle of the teeth may be obtained from a drawing as 
shown and explained in Fig. 10, or it may be calculated thus : 

„ r . „ circular pitch 

Tangent of angle *y = — : — ; — = *-? 

pitch circumference 

Tang - •* = 4.5X8 5 1416 =14^W = °-° 5805 

In Table No. 21 the corresponding angle is given as 3 
degrees, very nearly. 

The depth to which the gear should be cut is given in col- 
umn D as 0.515 inch. The gear is finished with a hob, as 
described below, which is allowed to cut until it touches the 
bottom of the spaces in the gear. The outside diameter of the 
hob should be larger than the outside diameter of the worm, in 
order that the teeth in the hob may reach the bottom of the 
spaces in the gear and leave clearance for the worm, and at the 
same time leave the gear tooth of the proper thickness on the 
pitch line. This increment is obtained in column &, Table No. 
50, and for %-inch pitch is 0.075 inch ; thus, the outside diameter 
of the hob is 0.075 inch larger than the outside diameter of the 
worm, or 4.977 + 0.075 = 5.052 inches. The angle of the finish- 
ing threading tool for both worm and hob is 14>^ degrees, 
making the angle of space 29°, as shown in Fig. 11. The clear- 
ance angle of the threading tool must be a little more than the 
angle of the thread, 



400 



WORMS AND WORM GEARS. 



The width of the threading tool at the point is given in 
Column b, Table No. 50, as 0.2325 inch. The depth of the space 
to be cut in the worm is given in Column Z>, as 0.515 inch. The 
diameter of the worm at the bottom of the thread will be : 
4.977 — 2 X 0.515 = 8.947 inches. 

The depth of the space to be cut in the hob is given in col- 
umn h in Table No. 50 as 0.5525 inch. 

The diameter of the hob at bottom of thread will be : 
5.052 — 2 X 0.5525 = 3.947 inches. 

Thus the only difference in size between the hob and the 
worm is in the outside diameter and in the depth of the cut. 
Both may be finished by the same tool, as the diameter at the 
bottom of the thread and the thickness of the teeth at the pitch 
line should be the same for both hob and worm. 



Elliptical Gear Wheels. 

Elliptical gear wheels are sometimes used in order to change 
a uniform rotary motion of one shaft to an alternately fast and 
show motion of the other. See Fig. 12. 




The pitch line is constructed and calculated the same as 
the circumference of an ellipse. (See page 189.) The gear is 
constructed involute the same as for spur gears. If the differ- 
ence between the minor and the major diameters is large it may 
be necessary to construct the teeth of different shapes at differ- 
ent places on the circumference ; in other words, the whole cir- 
cumference of the gear cannot be cut with the same cutter. A 
cutter of the same pitch, of course, but corresponding to a larger 
diameter of gear, must be used where the curve of the pitch line 
is less sharp. 

The centers of the shafts are in the foci of the ellipse. If 
two elliptical gear wheels, made from the same pattern, or cut 
together at the same time, on the same arbor, are to work together 
they must have an uneven number of teeth so that a space will 
be diametrically opposite a tooth, as will be seen from Fig. 12, 



SCREWS. 401 

SCREWS. 

"Pitch," "Inch Pitch" and "Lead" of Screws and Worms. 

The term " pitch of a screw," as commonly used, means its 
number of threads per inch, while the " inch pitch " is the dis- 
tance from the center of one thread to that of the next. For 
instance, a one-inch screw of standard thread is usually said to 
be an " eight pitch" screw, because it has eight threads per inch of 
length ; but it might more correctly be said to be a screw of 
>^-inch pitch, because it is J^-inch from the center of one thread 
to the center of the next. 

The " lead " of a worm or a screw means the advancement 
of the thread in one complete revolution : therefore, in a single- 
threaded screw, the inch pitch and the lead is the same thing, but 
in a double or triple-threaded screw the inch pitch and the lead are 
two different things. The " lead " in a double-threaded screw will 
be a distance equal to twice the distance from the center of one 
thread to the center of the next, but in a triple-threaded screw the 
lead is three times the distance from the center of one thread to 
the center of the next. 

Screw Cutting by the Engine Lathe. 

When the stud and the spindle run at the same speed 
( which they usually do) the ratio between the gears may always 
be obtained by simply ascertaining the ratio between the num- 
ber of threads per inch of the lead-screw and the screw to be 
cut. 

Example. 

The lead-screw on a lathe has four threads per inch and 
the screw to be cut has 11^ threads per inch ( one-inch pipe- 
thread). Find the gears to be used when the smallest change gear 
has 24 teeth and the gears advance by four teeth up to 96. 
The ratio of the number of threads per inch of the two screws 
is as 4 to 11^. 

As the smallest gear has 24 teeth and the gears all 
advance by four teeth, this ratio of the screws must be mul- 
tiplied by a number which is a multiple of 4 and which, at least, 
gives the smallest gear 24 teeth. For instance, multiply by 8 
and the result is 8 X 11 >£ = 92 teeth for the gear on the lead- 
screw ; S X 4 = 32 teeth for the gear on the stud. 

Cutting nultipIe=Threaded Screws or Nuts by the 
Engine Lathe. 

Calculate the change gears as if it was a single-threaded 
screw of the same lead. Cut one thread and move the tool the 
proper distance and cut the next thread. 



402 SCREWS. 

The most practical way to move the tool from one thread 
to another, when cutting double-threaded screws or nuts, is 
to select a gear for the stud or spindle of the lathe having a 
number of teeth which is divisible by two, and when one thread is 
cut make a chalk mark across a tooth in this gear onto the rim of 
the intermediate gear ; count half way around the gear on the stud 
and make a chalk mark across that tooth; drop the swing plate 
enough to separate the gears, pull the belt by hand until the oppo- 
site mark on the gear on the stud comes in position to match the 
chalk mark on the intermediate gear; clamp the swing plate again 
and the tool is in proper position to cut the second thread. 

When triple threads are to be cut, select a gear for the 
spindle or stud whose number of teeth is divisible by three, 
and in changing the tool from one thread to the next, only turn 
the lathe enough so that the gear on the stud moves one-third 
of one revolution. 

If, for any reason, it should be inconvenient to make this 
change by the gear on the stud, the change may be made by the 
lead-screw gear. The intermediate gear is first released from the 
gear on the lead-screw, which is then moved ahead the proper 
number of teeth, and again connected with the intermediate gear. 
The proper number of teeth to move the gear on the lead-screw 
is obtained by the following rule : 

Multiply the number of teeth in the gear on the lead-screw 
by the number of threads per inch of the lead-screw ; divide this 
product by the number of threads per inch of the screw to be 
cut, and the quotient is the number of teeth that the gear on the 
lead-screw must be moved ahead. 

Example. 

A square-threaded screw is to have J^-inch lead and triple 
thread. The lead-screw in the lathe has two threads per inch, 
and the gear on the lead screw has ninety-six teeth. How many 
teeth must the gear on the lead-screw be moved, when changing 
from one thread to the next ? 

Solution : 

A screw of >^-inch lead with triple thread has six threads 

2 X 96 
per inch, therefore the gear must be moved — ^ — = 32 teeth, 

in order to change the tool from one thread to the next. 



U. S. Standard Screws. 

Fig. 1 shows the shape of 
thread on United States stand- 
ard screws. The sides are 
straight and form an angle of 
sixty degrees, and the thread is 




SCREWS. 403 

flat at the top and bottom for a distance equal to one-eighth of 
t'le pitch, thus the depth of the thread is only three-fourths of 
a full, sharp thread. (See Fig. 1.) 

Fig. 2 shows the shape of the Whitworth (the English) sys- 
tem of thread. As compared =» ?, 

with the American system. ~""~j||^--552^/^. 0%\ 

the principal difference is in ^_ !%%$%>> /<Illlk J%%fy>>> 

the angle between the sides ** — ''^^^^k^^^^^^,^^^^^-, 
of the thread, which is fifty- ^11^1'— -'- -'£^^^^^^^^^^^ 
five degrees, and one-sixth of "" ^^^^^^^^^^^^^^^^^W 
the depth of the full, sharp ^^^^^ ^^^^ ^^^^ 

thread is made rounding at 2 

the top and bottom. There is also a difference in the pitch of 
a few sizes. 

The common Y-thread screws have the angle of thread of 
sixty* degrees, the same as the United States standard screws, 
but the thread is sharp at both top and bottom. This style of 
thread is rapidly, as it should be, going out of use. The prin- 
cipal disadvantages of this thread are that the screw has less 
tensile strength, and it is also very difficult to keep a sharp- 
pointed threading tool in order. 



Diameter of Screw at Bottom of Thread. 

The diameter of screws at the bottom of thread is obtained 
by the following formulas : — 

United States Standard Screws: 

1.299 



d = D - 

For V-threaded screws : 

d — D 

For Whitworth screws : 

d = D — 



71 



1.733 



71 



1.281 



n 

d '= Diameter of screw at bottom of thread. 
D = Outside diameter of screw. 
71 = Number of threads per inch. 
1.299 is constant for United States standard thread. 
1.733 is constant for sharp Y-thread. 
1.281 is constant for Whitworth thread. 



404 



SCREWS. 



TABLE No. 52.— Dimensions of Whitworth Screws. 



Diameter of 


Number of 


Diameter of 


Number of 


Diameter of 


Number of 


Screw in 


Threads per 


Screw in 


Tlireads per 


Screw in 


Threads per 


Inches. 


Inch. 


Inc'ies. 


Inch. 


Inches. 


Inch. 


% 


40 


Vi 


7 


3*A 


3X 


3 

16 


24 


1% 


6 


SU 


3 


X 


20 


iy 2 


6 


4 


3 


5 

16 


18 


iH 


5 


4X 


27/s 


H 


16 


1* 


5 


4^ 


27/ S 


16 


14 


1H 


4^ 


4X 


2H 


% 


12 


2 


±y 


5 


2% 


H 


11 


2% 


4 


0% 


2% 


% 


10 


2/ 2 


4 


oy 2 


2% 


H 


9 


m 


sy 2 


5^ 


2V 2 


1 


8 


3 


sy 2 


6 


2y 2 


1% 


7 


3K 


3X 







Diameter of Tap Drill. 

The diameter of the drill with which to drill for a tap is, if 
we want full thread in the nut, equal to the diameter of the 
screw at the bottom of the thread, and is, therefore, obtained by 
the same formulas. However, in practical work it is always ad- 
visable to use a drill a little larger than the diameter of the 
screw at the bottom of thread, because in threading wrought iron 
or steel the thread will swell out more or less, and a few thou- 
sandths must be allowed in the size when drilling the hole. In 
drilling holes for tapping cast-iron, a little larger drill is used, 
because it is unnecessary in a cast-iron nut to have exactly full 
thread. Table No. 53 gives sizes of drill for both wrought and 
cast-iron, which give good practical results for United States 
standard screws. 

Table No. 53 gives sizes of hexagon bolts and nuts. The 
size of the hexagon is equal to l l / 2 times the diameter of bolt + 
^-inch ; the thickness of head is equal to half the hexagon. 
The thickness of nut is equal to the diameter of the bolt. 
When heads and nuts are finished they are yVinch smaller. 

The table is calculated by the following formulas : 



d — D 



1.299 
n 



C = 1.155 A 



E=\A 



/= — X 

11 

B = 1AUA 



SCREWS. 405 

TABLE No. 53. — Dimensions of U. S. Standard Screws. 





"o 




.s 
















Finished 




5 




-3 


u 




Rough Bolt-Heads and 


Bolt- 









u 


s 

1— 1 




Nuts. 


Heads 







_j 




V 


6 






and Nuts 




« 






■a 

-a 






- S. 


VI O 


. ° 


i 


*J 


«j 


a 


. 


rt 


pq 


m tC 

— cs 


1-1 


4) 


- 


= A'i 


pq 



pq 


"0 

pq 
"o 


H 


g 

— — 


u 


■a 

c 

CS 

a 


.2 X 

■j . 


c 

O 

u 


C 
In 

O 


*o 






°Z 


I-* 


u . 


Ih 


— — 


"o 


r o 


■5 = 




v s 


in 

in 


0Q 

in 


■5 5 


In i 


V 




u 

E 


- g 


u 

u 

^3 


H 

CS 


V 

s. - 
r. ~ 


en - 

in es 


** to 

1/5 CS 




-• 
C 




in t; 

y. - 


y C 

= _ 


s 


= s 


cs 3 


g 




z - 


P 3 


O S 


r- 


O 


C = 


;•. 


cs 

3 


•2S 


- 
Q 


^ — 
--7- 


3 


CS 




5 * 


1- «J 

<3^ 




IS 
H 


r, ~ 

G 


3 

1 6 


z> 


</ 


J 1 


a 


« 


/ 


A 


1 B 


C £ | F 


X 0.185 A 


0.0269 


20 


0.0062 


X 


23 
3 2 


3 7 
64 


% 


X 


A 0.240 ^ 


0.0452 


18 


0.0069 


: 9 
3 2 


27 
32 


11 
16 


1 9 

6 4 


5 

1 6 


1 : 
32 


X 


Xs 0.294 i| 


0.0679 


16 


0.0078 


1 1 

1 6 


31 

3 2 


51 

64 


1 1 

3 2 


r ; s 


^ 


S 

1 6 


A 0.345 |3 


0.0935 


14 


0.0089 


2 5 

3 2 


l-6 7 4 


29 
3 2 


2 5 
64 


_7 

16 


2 3 

3 2 


X 


X 0.400 if 


0.1257 


13 


0.0096 


y* 


m 


1A 


1 6 


% 


1 3 

i -7 


7 
1 6 


9 'o.454 if 


0.1619 


12 


0.0104 


3 1 

32 


IX 


1^ 


31 

j A 


9 

1 6 


2 ■■• 

3 2 


X 


^ 0.507 if 


0.2019 


11 


0.0114 


h\ 


1^ 


itt 


\\ 


# 1 


9 

TB" 


X .0.620; j| 


0.3019 


10 


0.0125 


IX 


119 
x 64 


ItV 


H 


K lr 3 6 


1 1 
1 6 


Xs 0.731 H 


0.4197 


9 


0.0139 


1* 


_ 3 


lff 


23 

3 2 


7s 1^ 


1 3 
1 6 


1 0.838 II 


0.5515 


S 


0.0156 


m 


oi 9 
Z 64 


1^ 


1 3 
1 6 


1 It 9 . 


15 
1 6 


1^0.939 |i 


0.6925 


7 


0.0178 


m 


9 9 

*16 


2- 3 - 

z 3 2 


29 
3 2" 


iy 8 i 3 4 


ItV 


IX 1.065 1£ 


0.8892 


7 


0.0178 


2 


95 3 

■^64 


2tV 


1 


IX Hf 


1A 


l^i.iss.m 


1.0532 


6 


0.0208 


9 3 

-16 


q 3 

^32 


2H 


1A 


1^8 2^1 T \ 


lK1.284 1i| 


1.2928 


6 


0.0208 


2Xs 


3|f 


2 K 


i- 3 - 

16 


!^ 2^1-^ 


1# 1.889 1|J 


1.5153 


5X 


0.0227 


9_9_ 

-16 


m 


931 

Z 32 


1/2 


1# 2^1A 


1^ 1.490 


*x 


1.7437 


5 


0.0250 


2U 


d 64 


3-3- 

°1 6 


IH 


IX 2|iiH 


1^1.615' 


1H 


2.0485 


5 


0.0250 


m 


4-V 


Q 1 3 


115 

3 2 


M 2 ^ 1 il 


2 1.711 


123 


2.2993 


4X 


0.0278 


3X 


412 
^64 


3 3 L I 


1t 9 6 


2 3- 1 - in 


2X1.961 


Hi 
32 


3.0203 


4Xj 


0.0278 


3X 


4.^1 
^64 


^^ 


IK 


2 i/ 1 3JLl9_3_ 
A °16i^l6 


■2/ 2 2.175 


^ 


3.7154 


4 


0.0313 


w 


S3 1 
°64 


4.31 

^64 


Ht 


21/ 313 9-7- 

w /2 °16|-16 


2^ 2.425, 


2 fr 


4.6186 


4 


0.0313 


IX 


6 


4.29 
*3 2 


2>i 


23/ 4_3_ 911 

74 ^16-16 


3 2.629 


2 ^ 


5.42S4 


3X 


0.0357 


±Vs 


611 


Kll 

D 3 2 


9_5_ 

"16 


3 h% m 


3% 2.879' 


P 


6.5099 


3X 


0.0357 


5 


"fT 


^2 5 
°3 2" 


2^ 


ix m ,h% 


3>i 3.100 


fX 


7.5477 ; 


3X 


0.0384 


5# 


►73 9 
' 64 


6M 


911 
-16 


">y 3A3A 


3K 3.317 


;h 


8.6414 


3 


0.0417 


bU 


8^1 

^ b 4 


6*i 


2^ 


53/ Sll Oil 

°?4 T"6,°T6 


4 3.567 


!« 


9.9930 


3 


0.0417 


6X 




W 


3t^ 


1 I 6 T V 3if 


4# 3.798 


313 


11.3292 


2^ 


0.0435 


ox 


Q_3_ 

y l 6 


W 


3# 


*X 6 T y4 T \ 


4^ 4.028 


^ 


12.7366 


2Xj 


0.0455 


ey* 


»tf 


8 


»A 


1^,6414^ 


4K 4.255 


JA 


14.2197 


2# 


0.0476 


-% 


iox 


8^ 


3^ 


IX 


7 T 3 6 


4ii 

*1 6 


5 4.480 


i# 


15.7633 


2Xj 


0.0500 


-% 


ion 


81-f 


311 

16, 


5 


7A 


415 


5X|4.730 


i* 


17.5717 


2X 


0.0500 


8 


ll- 5 - 

11 ll) 


9X 


4 


5 x "Hsa 


o/ 2 4.953 


4ff 


19.2676 


2^8 ! 


0.0526 


sx 


11H 


OH' 


^A 1 


5 ^ S^5^ 


5X 5.203 


g*f 


21.2017 


2X 


0.0526 


sx 


12^ 


w? 


43^ 


5X 8fi 5H 


6 |5.423 


&!f 


23.0978 


2X 


0.0556 


9X 


12* 9 r i 


10H 


*16 


6 9 T V 5H 



406 



SCREWS. 



Note. — In finished work, the thickness of the head of the 
bolt and the nut is equal, and is He of an inch less than the 
diameter of the bolt. 

Columns B and C in Table No. 53 are very useful for mam- 
purposes : for instance, in selecting size of counter-bore when 
finishing castings, to give bearing for screw heads ; in turning 
blanks which are afterwards to be cut into square or hexagon 
heads, etc. 

Table No. 54.— Coupling Bolts and Nuts. 

(Hexagon). 
(All Dimensions in Inches) 





S. 


u 


~z — 


it 

- 


u 


rt 


l-l '_ 


- 


H - 


y* 


13 


Vs 


11 


H 


10 


ft 


9 


1 


8 


1# 


7 


1* 


1 



Dimensions of Head and Nut. 



Across 
the Flat. 



Across 

the 
Corner. 



Length 
of Head 
or Nut. 



I 1 / 

If* 

1- 

m 
9. 



m 

12JL 

x 3 2 

I'- 

9_3_ 
"3 2 

2 5. 

- 1 6 



5/8 

H 

1 




Table No. 55. — Round and Fillister Head Screws. 

(All Dimensions in Inches). 



Diameter of 
Screw. 


Number of 

Threads per 

Inch. 


Diameter of 
Head. 


Length of 
Head. 

1 


*= 


40 


3 


X 


3 

16 


24 


% 


_3 
16 


K 


20 


$i 


u 


5 

16 


18 


< 
T6 


5 
16 


- : : 


16 


9 
16 


H 


7 
T6" 


14 


Vs 


T§ 


% 


13 


H 


72 


9 
16 


12 


1 3 
16 


9 
16 


Vs 


11 


ft 


H 


H 


10 


1 


X 


ft 


9 


W 


% 


1 


8 


1* 


1 

! 




Round Head 
Cap Screw. 




Fillister Head 



SCREWS. 



407 



TABLE No. 56.— Dimensions of Hexagon and Square Head 

Cap Screws. 

(All Dimensions in Inches). 



Diameter of 
Screws. 


Threads per 
Inch. 


Hexagon Head. 


Square Head. 


Length of 
Head. 


Across 

the 

Flats. 


Across 

the 
Corners. 


Across 

the 

Flats. 


Across 

the 
Corners. 


A 

7 

16 

H 

9 
1 6 

H 

Y\ 

H 

1 


20 

18 

16 

14 

13 

12 

11 

10 

9 

8 

7 

7 


7 

1 6 

9 

T6- 
5/8 
% 

1 3 
16 

7/8 
1 

1# 

1* 

13/8 

l 1 ^ 


37 

64 

2 1 

32 
23 
32 
55 
64 
1 5 
16 

A 64 

l- 5 - 

i 32 

119 
-"-64 

l- 7 - 

A 16 

119 

147 

X 64 


3/8 

A 

9 

16 

1 1 
1 6 

H 

/8 

1% 

13/8 
1# 


1 7 

32 

5/8 

23 
32 
51 
64 
57 
64 
3 1 
32 

1* 

119 
x 32 
1 49 
X F4 

1« 

2/8 


5 

1 6 

/8 

7 
16 

% 

9 
1 6 

5/8 

/8 
1 

l 1 ^ 
1* 



Eye Bolts. 

It is very customary to weld an eye to a lag screw (see Fig. 
3) to use in handling heavy weights in shops. 

The following table (No. 57) gives the holding power of lag 
screws or eye bolts when screwed into spruce timber a little 
over the full length of thread. The suitable size of bit for the 
thread is also given in the table. 



TABLE No. 57- 



Fig 







Load at 




Diameter 


Diameter 


Which the 


Safe 


of Screw. 


of Bit. 


Screw 
Pulled Out. 


Load. 


1 inch 


S/i inch 


16,000 lbs. 


2,000 lbs 


/s " 


11 u 

16 


9,000 " 


1,125 " 


34 « 


5/8 " 


7,000 " 


875 " 


Ys " 


% " 


6,000 " 


750 " 


H " 


H " 


3,500 " 


437 " 


Y& " 


5 <( 
16 


1,900 " 


237 " 


Y " 


3 ti 
1 6 


700 " 


87 " 




4oS 



SCREWS. 



i ABLE No. 58. — Giving the Average Weight in Pouiil: 
per 100 Square Head Gimlet=Pointed Lag Screws. 



LENGTH 
in Inches. 


: 


5 
16 


- 1 
-■ 5 


^ 


V 


9 " 
16 


- : i 


*■ 


"A" 


1" 


^A 


m 


H 


7 


10 














2 


•3*2 


H 


8* 


12 


17 


24 


271 








' 2l A 


±\£ H 


9| 


14 


19 


26 


31 








3 


43/ 7^ 


11 


16 


21 


28 


34 


51 






3^ 


*% H 


12E 


18 


24 


31 


38 


55 






j. 


534 H 


14 


2'J 


26 


34 


42 


60 


85 


112 


±% 


G% IOi 


l-H 


22 


28 


37 


46 


65 91 


121 


5 


7 m 


17 


24 


32 


40 


~\ 1 1 


7'' 97 




0-, 


" : -- 12| 


1H 


26 


34 


43 


54 


76 


103 


140 


6 


8 131 


20 


28 


36 


46 


5 B 


81 


110 


15: 


6^ 






2H 


30 


B 


49 


62 


86 


117 


1 BO 


7 






23 


32 


41 


52 


G 5 


92 125 


170 


<^ 






24* 


34 


14 


5 5 


69 


97 132 




8 






26 


36 


47 


58 


( 3 


103 


140 


190 


sy 2 














i t 


108 




: 


9 














81 


113 


156 


2 ' 


9^ 














85 


118 


1 64 


22 


10 














89 


123 


172 


23> 


Size of 


K ,~ - s 


«B» 


-CI 


X 


* 




^ 




Kja 


Head in 


X X 


X 


X 


X 


X 


x 


* X 


X 


Inches. 








--_- 


«=-^ 




rr 55 tz<x 


^_ 



gimlet-pointed lag screw. 




Example. 

What is the weight of S lag screws 6" long and j 2 -mch 
in diameter ? 

Solution: 

Under the heading }4 -inch, in the line with 6 in the column 
of length, is the number 36. Thus. 100 lag screws of this size 
will average to weigh 36 pounds, and one such screw will weigh 
0-36 pound : 5 such screws will weigh 0.3 ,: i X 5 = 2.88 pounds. 



SCREWS. 



409 



French System of Standard Threads. 

In the French system of standard screws the thread has an 
angle of 60°, with flat top and bottom. (The French system is 
in this respect identically the same as the United States stand- 
ard thread.) The pitch and the diameter are given in millimeters 
(see Table No. 59). The form of thread is an equilateral tri- 
angle. The diameter of tap drill (or diameter of bolt or screw 
at bottom of the thread) is obtained in the following way : The 
height of an equilateral triangle is obtained by multiplying its 
base by 0.86603 (this number is sin. of 60°). Thus, assuming 
that the base = 1, and taking off one-eighth of the depth at top 
and bottom, that is reducing the depth of the thread one-fourth, 

,■-,.„, 0.86603 
the remaining depth will be 0.86603— 7 •= 0.64952 ; mul- 
tiplying this by 2, to allow for the depth of the thread on both 
sides of the screw, the constant will be 2 X 0.64952 = 1.29904, 
which for all practical purposes may be reduced to 1.3. Thus, 
when the pitch of the screw is 1 millimeter, the diameter of the 
screw at the bottom of the thread is 1.3 millimeters less than 
the outside diameter of the screw. Therefore, the diameter of 
the screw at the bottom of the thread may always be calculated 
by the simple rule: 

Multiply the pitch in millimeters by 1.3, and subtract the 
product from the outside diameter of the screw; the remainder 
is the diameter of the screw at the bottom of the thread, which 
is the same as the diameter of the tap drill, given in Table No. 59. 

TABLE No. 59.— French Standard Screws. 

(All Dimensions in Millimeters). 



Diameter of 


Pitch of 


Diameter of 


1 

Diameter of 


Pitch of 


Diameter of 


Screw. 


Screw. 


Tap Drill. 


Screw. 


Screw. 


Tap Drill. 


6 


1 


4.70 


36 


4 


30.80 


8 


1 


6.70 


42 


4.5 


36.15 


10 


1.5 


8.05 


48 


5 


41.50 


12 


1.5 


10.05 


56 


5.5 


48.85 


14 


2 


11.40 


64 


6 


56.20 


16 


2 


13.40 


72 


6.5 


63.55 


18 


2.5 


14.75 


80 


7 


70.90 


20 


2.5 


16.75 


88 


7.5 


78.25 


22 


2.5 


18.75 


96 


8 


85.60 


24 


3 


20.10 


106 


8.5 


94.95 


26 


3 


22.10 


116 


9 


104.30 


28 




24.10 


120 


9.5 


113.65 


30 


3.5 


25.45 


136 


10 


123.00 


32 


3.5 


27.45 


148 


10.5 


134.35 



4io 



SCREWS. 



German System of Standard Threads. 

In the German system of standard threads the angle is 53° 
7' 47". The reason for adopting such an odd angle is that the 
form of thread is a triangle, having its base equal to its height, 

and the top angle of such a triangle 
F |G - •*• is 53 c 7' 47". The thread in this 

system is also made flat at top and 
bottom equal to one-eighth of the 
pitch (see Fig. 4). The diameter 
of the screw at the bottom of the 
thread is. in this system, calculated 
by this rule : 
Multiply the pitch in millimeters by 1.5, subtract this pro- 
duct from the outside diameter of the screw and the remainder 
is the diameter of the screw at the bottom of the thread, which 
is the same as the diameter of the tap drill given in Table No. 60. 




TABLE No. 6o.— German Standard Screws. 

(All Dimensions in Millimeters). 













. 








o 




°53 


o 




o.-^ 


o 




°-£ 




o * 


So 


2 * 


o * 


Bq 


>1 

2 is 


"o * 


«P 


V u 


W 4) 


«j a 


V V 


4> 


u a 


V i) 


° V 


a o. 


S B 


■S 5 


c «< 


6 B 


-5 S 


£ rt 


S B 


f i B 


E rt 


.2« 


3-j: 


rt —i 


.2tfl 


2c/2 


« £-i 


.«C/3 


Sen 


- Ph 


P 


S 


Q 


R 


PL, 


P 


P 


Pm 


P 


1 


0.25 


0.625 


5 


0.8 


3.8 


20 


2.4 


16.4 


1.2 


0.25 


0.825 


5.5 


0.9 


4.15 


22 


2.8 


17.8 


1.4 


0.3 


0.95 


6 


1 


4.5 


24 


2.8 


19.S 


1.7 


0.35 


1.175 


7 


1.1 


5.35 


26 


3.2 


21.2 


2 


0.4 


1.4 


8 


1.2 


6.2 


28 


3.2 


23.2 


2.3 


0.4 


1.7 


9 


1.3 


7.05 


30 


3.6 


24.6 


2.6 


0.45 


1.125 


10 


1.4 


7.9 


32 


3.6 


26.6 


3 


0.5 


2.25 


12 


1.6 


9.6 


36 


4 


30 


3.5 


0.6 


2.6 


14 


1.8 


11.3 


40 


4.4 


33.4 


4 


0.7 


2.95 


16 


2 


13 








4.5 


0.75 


3.375 


18 


2.2 


14.7 









International Standard for Metric Screw Threads. 

An ^international standard for metric screw threads was dis- 
cussed at a congress which met for that purpose at Zurich, in 
October, 1898. The form of thread adopted is based on the 
Sellers thread, which it will be remembered has the shape of an 
equilateral triangle truncated one-eighth of its height at top and 
bottom, 



SCREWS. 



411 



To insure interchangeability, and to reduce the wear on taps 
and dies, the congress recommended that the bottom of the 
thread should be rounded off by any suitable curve, which should 
not deepen the cut more than an amount equal to Vie of the 
pitch beyond the standard Sellers type. The top of the thread 
is to be left flat, as in the Sellers system. The following stand- 
ard sizes and pitches were decided upon : 

TABLE No. 61. — International Standard Thread. 



Diameter in Milli- 


Pitch in Milli- 


Diameter in Milli- 


Pitch in Milli- 


meters. 


meters. 


meters. 


meters. 


6 and 7 


1 


30 and 33 


3.5 


8 and 9 


1.25 


36 " 39 


4 


10 and 11 


1.5 


42 " 45 


4.5 


12 


1.75 


48 " 52 


5 


14 and 16 


2 


56 " 60 


5.5 


18, 20 and 22 


2.5 


64 " 68 


6 


24 and 27 


3 


72 " 76 


6.5 



To Gear a Lathe to Cut Metric Thread when the 
Lead=Screw is in Inches. 

Use two intermediate gears, one having 100 teeth and the 
other 127 teeth, fasten these two gears together on the same hub, 
and gear the 100-tooth gear into the gear on the lead-screw 
and the 127-tooth gear into the gear on the stud (see Fig. 5). 

The lathe will then cut, practi- 
cally, one-half the number of threads 
per centimeter that it originally cut 
per inch with a common intermediate 
gear. For instance, the stud gear has 
24 teeth, the screw gear has 48 teeth 
and the lead-screw has four threads 
per inch ; the lathe will then, with a 
common intermediate gear, cut eight 
threads per inch, but by using such a 
double intermediate gear as is shown 
in Fig. 5 the lathe will cut four threads per centimeter, which is 
the same as one-fourth times 10 and equals 2 x / 2 millimeters pitch, 
which corresponds to a metric standard screw of IS millimeters 
diameter. 

To Calculate the Change Gear when Cutting metric 
Screws by an English Lead= Screw. 

Divide 20 by the pitch in millimeters, and the quotient is the 
corresponding number of threads per inch to which the lathe 
must be geared. 



Fig 




412 



SCREWS. 



Example 1. 

To gear a lathe in order to cut a metric standard screw 
24 millimeters in diameter and of three millimeters pitch, the 
lead-screw on the lathe having four threads per inch. 

Solution : 

Twenty divided by three gives 6%, therefore gear the lathe 
as if it was to cut 6% threads per inch with a common inter- 
mediate gear, and throw in the special intermediate gear as 
shown in the cut, and the lathe will cut a screw of three milli- 
meters pitch. The gearing is easily obtained, thus: 

The ratio between the screw gear and the stud gear is as 6% to 
4, which is the same as 20 to 12, or, reduced to its lowest terms, 5 
to 3. Hence, the gears may have any number of teeth providing 
the ratio is 5 to 8 ; for instance, multiplying by 9, 45 and 27 
could be used, or, multiplying by 10, 50 and 30 could be used, 
etc. 

TABLE No. 62.— How to Gear a Lathe when Cutting Metric Thread, 
Using Inch=Di\ ided Lead=5crew and Intermediate Gears, as 
Shown in Fig. 5. 



Screw to be Cut. 
(Pitch in Milli- 
meters). 








LEAD-SCREW ON 


LATHE. 


2 

Threads per 

Inch. 


3 

Threads per 
Inch. 


i 

Threa 
In 

u 

1-, w 


1 

ds per 
:h. 

w 


5 

Threads 
per Inch. 


6 

Threads 
per Inch. 


10 

Threads 
per Inch. 


3 be 
to 


v v 

co 


u 

_ M 
-a <u 

3 bC 

to 


CO 


20 


^» u 

V V 

80 


3 bX) 

24 


V V 

"-■ be 

80 


3 be 
CO 

20 


u I) 
u be 

CO 

40 


1 










24 


120 


1.5 


# 9 








24 


80 


30 


80 


36 


80 


80 


40 


2 


20 


100 


24 


80 


24 


60 


30 


60 


24 


40 


40 


40 


2.5 


20 


80 ! 


24 


64 


24 


48 


30 


48 


30 


40 


50 


40 


3 


24 


80 


27 


60 


24 


40 


30 


40 


36 


40 


60 


40 


3.5 


28 


80 


21 


10 


28 


40 


35 


40 


42 


40 


70 


40 


4 


32 


80 


24 


40 


32 


40 


40 


40 


48 


40 


80 


40 


4.5 


27 


60 


27 


40 


36 


40 


45 


40 


54 


40 






5 


30 


60 


30 


40 


40 


40 


50 


40 


60 


40 






5.5 


33 


60 1 


33 


40 


44 


40 


55 


40 


66 


40 






6 


36 


60 1 


36 


40 


48 


40 


60 


40 


72 


40 






6.5 


39 


60 


39 


40 


52 


40 


65 


40 










7 


28 


40 ! 


42 


40 


56 


40 


70 


40 










7.5 


30 


40 


45 


40 


60 


40 














8 


32 


40 


48 


40 


64. 


40 














8.5 


34 


40 


51 


40 


















9 


36 


40 


54 


40 


















9.5 


38 


40 


57 


40 


















10 


40 


40 


60 


40 


















10.5 


42 


40 






















11 


44 


40 






















11.5 


46 


40 






















12 

i 


48 


40 | 























NOTES ON HYDRAULICS. 4 1 3 

NOTES ON HYDRAULICS. 

Hydraulics is the branch of engineering treating on fluid in 
motion, especially of water, its action in rivers, canals and pipes, 
the work of machinery for raising water, the work of water as 
a prime mover, etc. 

Pressure of Fluid in a Vessel. 

When fluid is kept in a vessel the pressure will vary directly 
as the perpendicular height, independent of the shape of the 
vessel. For water, the pressure is 0.434 pounds per square inch, 
when measured one foot under the surface. The pressure in 
pounds per square inch may, therefore, always be obtained by 
multiplying the head by 0.434. The head corresponding to a 
given pressure is obtained by either dividing by 0.434 or multi- 
plying by 2.304. 

Example. 

What head corresponds to a pressure of 80 pounds per 
square inch ? 
Solution : 

80 X 2.304 = 184 feet. 

Velocity of Efflux. 

The velocity of the efflux from a hole in a vessel will vary 
directly as the square root of the vertical distance between the 
hole and the surface of the water. For instance, if an opening 
is made in a vessel four feet, and another 25 feet, below the sur- 
face of the water, and the vessel is kept full, the theoretical 
velocity of the efflux will be nearly 16 feet and 40 feet per second 
respectively, friction not considered; or, in other words, the 
velocity will be as 2 to 5, because V 4 = 2 and V25 = 5. 

The velocity of efflux in feet per second may always be 
calculated theoretically by the formula : 

v — 8.02 X V7T . 

Constant 8.02 is \S-2g = \A>4.4, and v = velocity of efflux. 
h = Head in feet. 

Table No. 63 gives the theoretical velocity of efflux and the 
static pressure corresponding to different heads, and is calcu- 
lated by the following formulas : 

h = ~ v = V/zX 64.4 v = s/ P X 2.3 X 64.4 

v = V P X 148 P — hX 0.434 h = P X 2.3 



4i4 



NOTES ON HYDRAULICS. 



TABLE No. 63.— Head, Pressure, and Velocity of Efflux 

of Water. 



Head in 
Feet. 


Pressure in 

Pounds per 

Square Inch. 


Velocity in 
Feet per 
Second. 


Head in 
Feet. 


Pressure in 

Pounds per 
Square Inch. 


Velocity in 
Feet per 
Second. 


h 


P 


V 


h 


P 


V 


0.1 


0.0434 


2.54 


19 


8.246 


35 


0.2 


0.0868 


3.59 


20 


8.68 


35.9 


0.25 


0.1082 


4.01 


25 


10.82 


40.1 


0.3 


0.1302 


4.39 


30 


13.02 


43 


0.4 


0.1736 


5.07 


35 


15.19 


47.4 


0.5 


0.217 


5.67 


40 


17.36 


50.7 


0.6 


0.2604 


6.22 


45 


19.53 


53.8 


0.7 


0.3038 


6.71 


50 


21.7 


56.7 


0.75 


0.3255 


6.95 


55 


23.87 


59.5 


0.8 


0.3472 


7.18 


60 


26.04 


62.1 


0.9 


0.3906 


7.61 


65 


28.21 


64.7 


1 


0.434 


8.02 


70 


30.38 


67.1 


1.25 


0.5425 


8.95 


75 


32.55 


69.5 


1.5 


0.651 


9.83 


80 


34.72 


71.8 


1.75 


0.7595 


10.6 


85 


36.89 


73.9 


2 


0.868 


11.4 


90 


39.06 


76.1 


2.25 


0.9735 


12 


95 


41.23 


78.2 


2.5 


1.082 


12.6 


100 


43.4 


80.2 


2.75 


1.1905 


13.3 


110 


47.74 


84.2 


3 


1.302 


13.9 


120 


52.08 


87.68 


3.25 


1.4102 


14.4 


130 


56.78 


91.5 


3.5 


1.519 


15 


140 


61.06 


94.7 


3.75 


1.6375 


15.5 


150 


65.1 


98.3 


4 


1.736 


16 


160 


69.44 


101.2 


4.25 


1.8445 


16.5 


170 


73.78 


104.5 


4.5 


1.953 


17 


180 


78.12 


107.2 


4.75 


2.0615 


17.5 


190 


82.46 


110.4 


5 


2.17 


17.9 


200 


86.8 . 


113.5 


6 


2.604 


19.6 


225 


97.35 


120 


7 


3.038 


21.2 


250 


108.2 


126 


8 


3.472 


22.8 


275 


119.05 


133 


9 


3.906 


24.1 


300 


130.2 


139 


10 


4.34 


25.4 


325 


141.05 


144 


11 


4.774 


26.6 


350 


151.9 


150 


12 


5.208 


27.8 


375 


163.75 


155 


13 


5.678 


28.9 


400 


173.6 


160 


14 


6.106 


30 


425 


184.45 


165 


15 


6.51 


31.1 


450 


195.3 


170 


16 


6.944 


32.1 


475 


206.15 


174 


17 


7.378 


33.1 


500 


217 


179 


18 


7.812 


34 


550 


238.7 


188 • 



NOTES ON HYDRAULICS. 415 

Velocity of Water in Pipes. 

The theoretical velocity of water discharged from a pipe is 
calculated by the same formula as is used in calculating veloci- 
ties of falling bodies. ( See page 277). 

v = V 2 g h 

v = Theoretical velocity of efflux per second. 
h = Head. 
2 g = 64.4 if v and h are reckoned in feet. 
2 g = 19.64 if v and h are reckoned in meters. 
If the water, besides the pressure due to the head, is also 
acted upon by some additional pressure, for instance, stea«t, 
the theoretical velocity of the discharge is obtained by the 
formula, 



\ ^ * ^ -r QAU J 



P = Pressure in pounds per square inch. 

The constant 0.434 is used because a column of water one 
foot high will exert a pressure of 0.434 pounds per square inch ; 
thus, by dividing by 0.434, we actually convert the pressure into 
its corresponding head in feet. 

All other quantities in this formula are, of course, taken in 
English units. 

Note. — By head is always meant the vertical height 
in feet, or its equivalent in pressure expressed in feet. Table 
No. 63 gives the theoretical velocity of the discharge and the 
pressure corresponding to different heads. 

The theoretical velocity is never obtained in practice, be- 
cause part of the total head is used to overcome the resistance 
at the entrance of the pipe, and part is used to overcome 
the frictional resistance to the flow of the water in the pipe. 
Thus, only a part of the total head is left to give 
velocity to the water, therefore the velocity of the water at dis- 
charge will only be what is due to the velocity head, after deduc- 
tions are made for resistance at the entrance and for friction in 
the pipes. In short pipes, the resistance at the entrance to the 
pipe is comparatively the larger loss, but in long pipes the fric- 
tional resistance is the larger. 

When both the resistance at the entrance and the friction 
in the pipe are considered the formula will be : 



«=Vi 






v = Velocity of discharge in feet per second. 
2 g = 64.4 

L ■= Length of pipe in feet. 

d = Diameter of pipe in feet. 

f — Coefficient of friction, which is obtained from experi- 
ments, and will vary according to conditions, from 0.01 to 0.05. 
It is usually in approximate calculations taken as 0.025. 



41 6 notes on hydraulics. 

Example. 

Find the velocity of discharge from a pipe six inches in diam- 
eter. The head is 16 feet and the length of the pipe is 100 
i.ee% and coefficient of friction Out _ ' 

Solution : 

[Note. 6 inches = 0.5 foot.) 



64.4 X 16 



~^ 1.5 +0.025 X-^ 



' 6.5 

- = 12j6 feet per second. 

In Table No. 64 the quantity of water discharged per min- 
ute by a pipe six inches in diameter, when the velocity is one foot 
per second, is ss.14 gallons. Thus, the quantity of water deliv- 
ered when the velocity is 1. feet per second, is 12.6 X S8. 14 = 
1110.6 gallons per minute. 

\\~hen the length of the pipe is more than 4.000 diameters 

tae velocity of the water may be calculated by the formula, 



and the quantity is obtained by multiplying the velocity by the 
constants given in Table No. 64. 

Example. 

Find the velocity of efflux from a water pipe of three 
inches diameter and 1200 feet long, having a head of six feet, 
assuming coefficient of friction as OjOS : 

Solution: 



64.4 X 6 

= 1.79 teet per second 



* = V..*. 5 x^ 



0.25 



Discharge in gallons per minute : 

q = 22.0S < 1.79 = 39.4 gallons per minute. 



NOTES ON HYDRAULICS. 



417 



TABLE No. 64.— Quantity of Water Discharged Through 

Pipes in One Hinute, when Velocity of Efflux 

is One Foot per Second. 



"o 

>-t . 

U IT. 

2JS 

is 
s c 

i« 

n 0. 
Sffi 

c 
p— 1 


Internal Diameter of 
Pipe in Feet. 


Internal Area of Pipe 
in Square Feet. 


Discharge in Cubic 
Feet per Minute at 
Velocity of One 
Foot per Second. 


Discharge in Gallons 
per Minute at a 
Velocity of One 
Foot per Second. 


"0 
■- 

si 

to - 

as 

c 


u 

V 
D . 

£ 4> 
TO W 

as 

a 


u 

O V 

2 £ 

< TO 
— 3 
TO J 

czn 

a s 
1— 1 


Discharge in Cubic 
Feet per Minute at 
a Velocity of One 
Foot per Second. 


Discharge in Gals. pr. 
Minute at a Veloc- 
ity of 1 Foot pr. Sec. 


% 0.0104 0.00008 


0.0048 


0.036 


20 


1.6666 


2.182 


130.90 


979 


% 0.0208 0.00033 


0.0198 


0.150 


| 21 


1.7500 


2.405 


144.32 


1079 


% 0.0312 0.00076 


0.0456 


0.342 


22 


1.8338 


2.640 


158.39 


1185 


X 0.0416 0.00136 


0.0816 


0.612 


J 23 


1.9166 


2.885 


173.11 


1294 


2£ 0.0624 0.00306 


0.1836 


1.380 


24 


2.000 


3.142 


188.50 


1410 


1 10.0833 0.00545 


0.3272 


2.448 


25 


2.0833 


3.409 


204.53 


1530 


li- 0.1042 0.00852 


0.5094 


3.828 


26 


2.1667 


3.687 


221.22 


1655 


1* 


0.1250 0.01227 


0.7362 


5.508 


27 


2.2500 


3.976 


238.56 


1784 


2 


0.1667 0.0218 


1.309 


9.792 


128 


2.3333 


4.276 


256.56 


1919 


2^ 0.2083 0.0341 


2.045 


15.30 


29 


2.4166 


4.578 


275.22 


2058 


3 10.2500 0.0491 


2.945 


22.03 


30 


2.5000 


4.909 


294.52 


2203 


3i 0.2911 0.0668 


4.008 


29.99 


31 


2.5822 


5.241 


314.49 


2352 


4" 0.3333 0.0873 


5.238 


39.17 


32 


2.6667 


5.585 


335.10 


2506 


4i 0.3750 0.1104 


6.626 


49.58 


|33 


2.7500 


5.939 


356.37 


2666 


5 0.4166 0.1364 


8.181 


61.20 


34 


2.8333 


6.305 


378.30 


2829 


6 0.5000 0.1963 


11.781 


88.14 


35 


2.9166 


6.6S1 


400.88 


2999 


7 0.5822 


0.2673 


16.035 


119.9 


;36 


3.0000 


7.069 


424.14 


3173 


8 0.6667 


0.3491 


20.914 


156.7 


37 


3.0S33 


7.467 


448.02 


3351 


9 0.7500 


0.4418 


26.507 


198.3 


38 


3.1667 


7.876 


472.56 


3535 


10 0.8333 


0.5454 


32.725 


244.8 


39 


3.2500 


8.296 


497.75 


3724 


11 


0.9166 


0.6597 


39.597 


296.2 


40 


3.3333 


8.727 


523.60 


3918, 


12 


1.0000 


0.7854 


47.124 


352.5 


41 


3.4166 


9.168 


550.11 


4115 


13 1.0833 


0.9218 


55.305 


413.7 


42 


3.5000 


9.621 


577.27 


4318 


14 


1.1667 


1.069 


64.141 


479.8 


43 


3.5822 


10.085 


605.09 


4526 


15 


1.2500 


1.227 


73.631 


550.8 


44 


3.6667 


10.559 


633.56 


4739 


16 


1.3333 


1.396 


83.776 


626.4 


45 


3.7500 


11.045 


662.68 


4961 


17 1.4166 


1.576 


94.575 


707.4 


46 


3.8333 


11.541 


692.46 


5180 


18 1.5000 


1.768 


106.03 


793.2 


47 


3.9166 


12.048 


722.90 


5408 


19 1 1.5822 


1.969 


118.14 


883.8 


48 


4 


12.566 


753.98 


5640 





4i3 



sorcx= :n stia::. 



NOTES ON STEAM. 

When vrater is heated amverted iato steam of atmos- 

pheric pressure, one cubic foot of water will make 1646 cubic 
feet of steam. (The common expression that " a cubic inch of 
water makes a cubic foot of steam" is not strictly correct, as a 

cucic :::: zzizains ITi'S zidziz iizies 

The sp - ::d : rravirr : : steara ;.: atari : spaer: : press ire. vdaea 



compared witb 

The weig] 
pressure will, t 
any other presi 
in Table Nc 

Saturated 
p :ii: — ki:h : : 
act :::: :: ': e 
that da e steara 
sarirated ~ddz 
zcivea a res scire 
szrcri is in ::: 



a: atrazspaeriz 
i ->S p zaads. A: 
szeaal is rivea 



: jiling 

Lit IItS 

:: aaeaa 
a: i: is 



more ~azer ".i_ ;e evapzratei. aaa it me v nance :s ceit :::.- 
staat. as ia :-. steam ccder, both die iressire aaa remaerainre 
will ia : r e a s e s i : a eously . 

Hip: : ■::. .. ■>■■: ::■:.: ~<: is s:eari die iressire :f — hich rrtai 
exceeds the pressure of the atmosphere. 

Lon pressure steam, is steam die pressure of which is less 
daaa die amicsphere. aaa als: steara havzic: a iressire eriai 
to, or not greatly above, the atmospheric pressure. 

Wet steam is steam which contains water held in suspen- 
sion mechanicallv. 

Dry steam is steam which does not contain water held in 
suspension raezhaaizady. 

Super-heated steam is steam which is heated to a tempera- 
rire diaier daaa die ': ziiiii p :da: zomespondmg- to its pressure. 
It zciizz exist ii contact with water, nor contain water, and 
resercbies a percezc ccas. Verdzal bzrers ~.dth razes ticrzizdc 
the steam spa ze sizh as dae idimcm.Lc b cider cdve sii zcddv sap er- 
heated steam ; but if steam is to be super-heated to any consid- 
erable extent it must be passed through a super-heater, which 
usually is in the form of a coil of pipes subjected to the hot 
cases ia dee uptake frzai die z :der 

The sensible heat of steam is tie temperature which can 
be measured by a thermometer. 

The latent Jieat of steam is that heat which is absorbed 
when water of any given temperature is changed into steam of 
the same temperature. 

When water is evaporated under pressure the sensible heat 
will increase and the latent heat will decrease. For instance, 
at atmospheric pressure the sensible heat is 212 decrees, and the 
latent heat of evaporation is 9*56 B. T. U.« but at 100 pounds 



NOTES ON STEAM. 



419 



absolute pressure the sensible heat is 327.9 degrees, while the 
latent heat of evaporation is only 883.1 B. T. U. (See steam 
table, No. 65.) 



TABLE No 


. 65.— Properties 


of Saturated Steam. 


Absolute Pres- 
sure in Pounds 
per Square 
Inch. 




V ~ V 

§•«•§& 


• ~ a w (it 


c .-* 

rt • - 1— ' -d . 


V 

c »~- 

V 3 

■I" °^ 
u 


C l> « 

— a • 
j-> g 

►> o,n 

t>CL, 3 


*0 _ Oca . 

r s s r 8ts s 

M-i •*- fa u u 
3 w :>Q 


1 


102.1 


1112.5 


1042.9 


330.36 


0.0030 


20800 


2 


126.3 


1119.7 


1025.8 


172.80 


0.0058 


10760 


3 


141.6 


1124.6 


1014 


117.52 


0.0085 


7344 


4 


153.1 


1128.1 


1006.8 


89.36 


0.0112 


5573 


5 


162.3 


1130.9 


1000.3 


72.80 


0.0138 


4524 


6 


170.2 


1133.3 


995 


61.52 


0.0163 


3813 


7 


176.9 


1135.3 


990 


52.62 


0.0189 


3298 


8 


182.9 


1137.2 


985.7 


46.66 


0.0214 


2909 


9 


188.3 


1138.8 


982.4 


41.79 


0.0239 


2604 


10 


193.3 


1140.3 


978.4 


37.84 


0.0264 


2358 


11 


197.8 


1141.7 


975.3 


34.63 


0.0289 


2157 


12 


202 


1143 


972.2 


31.88 


0.0314 


1986 


13 


205.9 


1144.2 


970 


29.57 


0.0338 


1842 


14 


209.6 


1145.3 


968 


27.61 


0.0362 


1720 


14.7 


212 


1146.1 


966 


26.36 


0.0380 


1646 


15 


213.1 


1146.4 


964.3 


25.85 


0.0387 


1610 


16 


216.3 


1147.7 


962.6 


24.32 


0.0411 


1515 


17 


219.6 


1148.3 


960.4 


22.96 


0.0435 


1431 


18 


222.4 


1149.2 


957.7 


21.78 


0.0459 


1357 


19 


225.3 


1150.1 


956.3 


20.70 


0.0483 


1290 


20 


228 


1150.9 


952.8 


19.72 


0.0507 


1229 


25 


240 


1154.6 


945.3 


15.99 


0.0625 


996 


30 


250 


1157.8 


937.9 


13.46 


0.0743 


838 


35 


259.3 


1160.5 


931.6 


11.65 


0.0858 


726 


40 


267.3 


1162.9 


926 


10.27 


0.0974 


640 


45 


274.4 


1165.1 


920.9 


9.18 


0.1089 


572 


50 


281 


1167.1 


916.3 


8.11 


0.1202 


518 


55 


287.1 


1169 


912 


7.61 


0.1314 


474 


60 


292.7 


1170.7 


908 


7.01 


0.1425 


437 


65 


298 


1172.3 


904,2 


6.49 


1538 


405 


70 


302.9 


1173.8 


900.8 


6.07 


0.1648 


378 


75 


307.5 


1175.2 


897.5 


5.68 


0.1759 


353 


80 


312 


1176.5 


894.3 


5.35 


0.1869 


333 


85 


316.1 


1177.9 


891.4 


5.05 


0.1980 


314 


90 


320.2 


1179.1 


888.5 


4.79 


0.2089 


298 


95 


324 


1180.3 


885.8 


4.55 


0.2198 


283 


100 


327.9 


1181.4 


883.1 


4.33 


0.2307 


270 


105 


331.3 


1182.4 


881.7 


4.14 


0.2414 


257 



420 



NOTES ON STEAM. 



TABLE No. 65. — (Continued). 



te Pres- 
Pounds 
quare 
ich. 


erature 
oiling 
nt in 
ees F. 


Heat in 
U. per 

Df Steam 
Vater at 
;rees F. 


Heat of 
iration 
'. U. per 
nd of 
am. 


me in 
Feet per 
nd of 
am. 


jht in 
ds per 
Foot of 
am. 


Feet of 
from 1 
Foot of 
rat 62 
ees F. 


Absolu 

sure in 

per S 

In 


Temp 

or B 

Poi 

Degr 


Total 
B. T. 
Pound 1 
from V 
32 Det 


Latent 

Evap( 

in B. 1 

Pou: 

Ste 


Volu 

Cubic ] 

Pou 

Ste 


Wei? 

Poun 

Cubic 

Ste 


Cubic 

Steam 

Cubic 

Wate 

Degr 


110 


334.6 


1183.5 


878.3 


3.97 


0.2521 


247 


115 


338 


1184.5 


875.9 


3.80 


0.2628 


237 


120 


341.1 


1185.4 


873.7 


3.65 


0.2738 


227 


125 


344.2 


1186.4 


871.5 


3.51 


0.2845 


219 


130 


347.2 


1187.3 


869.4 


3.38 


0.2955 


211 


135 


350.1 


1188.2 


867.4 


3.27 


0.3060 


203 


140 


352.9 


1189 


865.4 


3.16 


0.3162 


197 


145 


355.6 


1189.9 


863.5 


3.06 


0.3273 


190 


150 


358.3 


1190.7 


861.5 


2.96 


0.3377 


184 


160 


363.4 


1192.2 


857.9 


2.79 


0.3590 


174 


170 


368.2 


1193.7 


854.5 


2.63 


0.3798 


164 


180 


372.9 


1195.1 


851.3 


2.49 


0.4009 


155 


190 


377.5 


1196.5 


848 


2.37 


0.4222 


148 


200 


381.7 


1197.8 


845 


2.26 


0.4431 


141 



In the preceding table the first column gives the absolute 
pressure, which is gage pressure plus 14.7 pounds, or, for 
ordinary practice, reckon as 15 pounds. For instance, when 
the gage pressure is 80 pounds per square inch, the correspond- 
ing absolute pressure is, for all practical purposes, 95 pounds 
per square inch, and the corresponding temperature is given in 
the second column in the table to be 3^4 degrees Fahr. 

The total number of British thermal units (B. T. U.) re- 
quired to convert each pound of water from 32 degrees Fahr. 
into steam of any given pressure is given in the third column. 
For instance, each pound of water of 32 degrees converted 
into steam of 95 pounds per square inch absolute pressure has 
received 1180.3 B. T. U. 

The fourth column gives the number of British thermal 
units (B. T. U.) of heat required to change one pound of water 
of the temperature given in the second column into steam of 
the same temperature ; which also is the number of heat 
units given up by one pound of steam when it is condensed to 
water of the same temperature as the temperature of the steam 
with which it is in contact. For instance, the table gives the 
latent heat of evaporation of steam at 95 pounds absolute pres- 
sure to be 885.8 B. T. U.; therefore, if steam of 95 pounds pres- 
sure per square inch is condensing into water in a steam pipe 
where steam and water are in contact, so that the temperature 
cannot drop below that due to the pressure, and the pressure is 



NOTES ON STEAM. 42 I 

maintained at 95 pounds per square inch, the temperature of 
the water from the condensed steam will be 324 degrees, the 
yame as the temperature of the steam, but each pound of steam 
as it is condensing will give out SS5.8 British thermal units of 
heat. 

The fifth column gives the number of cubic feet of satu- 
rated steam which will weigh one pound at the given pressure 
and temperature. The sixth column gives the weight of one 
cubic foot of saturated steam of corresponding given tempera- 
ture. For instance, one cubic foot of steam at 95 pounds per 
square inch absolute pressure will weigh 0.2198 pounds, and 100 
cubic feet of steam of 95 pounds per square inch absolute pres- 
sure will weigh 100 X 0.2198 = 21.98 pounds. In other words, it 
will take 0.2198 pounds of water to give one cubic foot of steam 
at 95 pounds absolute pressure, and it will require 21.9S pounds 
of water to make 100 cubic feet of steam of 95 pounds absolute 
pressure. 

The seventh or last column gives the relative volume of 
steam at the given pressure as compared with water at 32 
degrees F. For instance, one cubic foot of water will give 
1646 cubic feet of steam at atmospheric pressure, but one cubic 
foot of water gives only 219 cubic feet of steam at 125 pounds 
absolute pressure. 

Steam Heating. 

In the ordinary practice of heating buildings by direct 
radiation the quantity of heat given off by the radiators or steam 
pipes will vary from 1% to 3 heat units per hour per square foot 
of radiating surface for each degree of difference in tempera- 
ture; an average of from 2 to 2% is a fair estimate. 

One pound of steam at about atmospheric pressure contains 
1146 heat units, and if the temperature in the room is to be 
maintained at 70°, while the temperature of the pipes is 
212°, the difference in temperature will be 142 degrees. Multi- 
plying this by 2%, the emission of heat will be 229>^ heat units 
per hour per square foot of radiating surface. Dividing 229 j^ 
by 1146 gives 0.2 pounds of steam condensed per hour, per square 
foot of radiating surface. From this may be estimated the re- 
quired size of boiler, as the boiler must always be capable of 
generating as much steam as the radiators are condensing. A 
rule frequently given is to have one square foot of heating sur- 
face in the boiler for every 8 to 10 square feet of radiating sur- 
face and one square foot of grate surface for every 350 to 500 
square feet of radiating surface. 

One pound of coal is required per hour per 30 to 40 square 
feet of radiating surface. 

When steam is used for heating dwelling-houses, one square 
foot of radiating surface is required per 40 to 80 cubic feet of 
space, according to location, number of windows, etc. As a 



it : y :~ l~ :.' : . _ -i: 



I'tlT.*!. M_r !1: M MM M«M ~Z MI MMMT = ZZTZ 1 ! r 15 

far heating 40 to (SO cnbic feet of air m outer or front 

izi sj :: :.i ;_ :;:--- : .^:;;>:::- 7 t : _: ~mz rzlt m~ 
t isei i-r i --It ::: M:::i- ::m:::.: Jim - c mm : :«m :: 
M.f . i- mm - mm m r - ~ : :i" : - zzz: izm • :.: s :.;;:::: :: 
- : i : t m : r_ . z z = r : : • : - mm : m : m " ' : : 1 1 it. : . : : - - : : : 
5; i:t m - -"- :m ::..•:• mi mmi - :.- M :: . ■ • ; _: : :-: 
; : - - m m m m t i m : Lit ^ t 1 tmm 1 — ~ ms 

I" Z±Z.Z ZZ : _ : t t rMMM__T ^sei. 

mm :-t : :zziz i±M :: mmm :: ~e 



\"iu: 



- - \\t- f - -M-.- 



:1";_ 



Quantity of Water Required to Make awy guim 
5:eiM 2: 2-" 



:: mmm : z- m : . : m«m :; 

m m r " r mm : :z- ; _ : .; 
iImml m 7::.: ; .! 

L= : immm-c :t ~ __- :. ;-.m 
::.Jt: :j :_„t = rl: : : 



- : — 1: 
1 : 1 ■ ::-ii 

5mm£m 

: -t :_i 
±r m:.t m 

_ - t.M : 

\ z- zzzz 
- - : •: : i?: :: " 



NOtES ON 9fEAM. 423 

taken as B T % pounds. Therefore 69.21 pounds divided by 62.5 
gives 1.1 cubic feet, or 69.21 pounds divided by 8.3 gives 8.34 
gallons. 

At atmospheric pressure one cubic foot of steam has nearly 
the weight of one cubic inch of water, and the weight increases 
very nearly as the pressure ; therefore, for an approximate estima- 
tion, if no steam tables are at hand, it is well to remember the 
rule : 

Multiply the number of cubic feet of steam by the absolute 
pressure in atmospheres, and the product is the number of cubic 
inches of water required to give the steam. 

Note. — In all such calculations for practical purposes, a 
liberal allowance must be made for loss and leakage. 

Weight of Water Required to Condense One 
Pound of Steam. 

The following formula gives the theoretical amount of water 
required to condense one pound of steam : 

= /r +82-* 

fa— h 

W = Weight of water required per pound of steam con- 
densed. 
H = Number of heat units above 32° in one pound of steam 
at the pressure of exhaust. This temperature is 
obtained from Table No. 65. 
t\ = Temperature of water when entering the condenser. 
fa = Temperature of water when leaving the condenser. 
h = Temperature of the condensed steam when leaving 
the condenser and entering the air-pump. 

Example. 

Steam of four pounds absolute pressure is exhausted into a 
surface condenser. The temperature of the condensed steam 
when leaving the condenser and entering the air-pump is 120°. 

The temperature of the cold water when entering the con- 
denser is 65°. 

The temperature when leaving the condenser is 105°. 
How many pounds of condensing water is needed per pound of 
steam condensed ? 

Note. — In the steam table, page 419, the total number of 
heat units above 32° per pound of steam of four pounds absolute 
pressure is given as 1128. 

Solution : 

1128 + 32 — 120 



W = 



105 — 65 



W = = 26 pounds of water per pound of steam. 



424 NOTES ON STEAM. 

In a jet condenser the steam and the water are mixed to- 
gether, and, therefore, the condensed steam and the water when 
leaving the condenser are of equal temperature, and the formula 
will change to 



to — Temperature of mixture. 

t\ = Temperature of water when entering condenser. 

The other letters have the same meaning as in the previous 
formula. 

Example. 

Steam of three pounds absolute pressure in exhausted into 
a jet condenser. The temperature of the cold water entering is 
60°. The temperature of the mixture leaving the condenser is 
110°. How many pounds of water are needed per pound of 
steam condensed ? 

Note. — In the steam table, page 419, the total number of 
heat units above 32 c per pound of steam of three pounds pres- 
sure is given as 1124.6 or, for convenience, say 1125. 

Solution : 

1125 + 32 — 110 



W = 



110 — 00 



W ' = ,^ = 20.1 pounds. 
oO 

Weight of Steam Required to Boil Water. 

An approximate rule is to allow that one pound of steam is 
condensed for every five pounds of water to be heated to the 
boiling point. 

It does not make much difference about the pressure of the 
steam, as long as it is a few pounds above atmospheric pres- 
sure ; for instance, one pound of steam at 10 pounds gage pres- 
sure when condensed into water at 212° will give up 973 heat 
units, and steam of 100 pounds gage pressure will give up 1003 
heat units — a difference of only 30 heat units in steam of 10 
pounds gage pressure and steam of 100 pounds gage pressure. 

More correctly, the weight of steam required to boil one 
pound of water at 212 c may be calculated by the formula, 
_ 212 — h 
X ~ H— 180 
And the weight of steam required to heat one pound of water to 
any temperature is obtained by the formula, 

_ ti— t\ 
X ~ H + 32 — h 



NOTES ON STEAM. 425 

x = Weight of steam required. 

H = Number of heat units above 32° in one pound of steam, 
as given in Table No. 65. 

t\ — Temperature of water before heating. 

/2 = Temperature of water after heating. 



Expansion of Steam in Steam Engines. 

When steam is expanded without doing work and prac- 
tically without losing heat by radiation, it will become super- 
heated, but if it is doing work, as in a steam engine, it will lose 
heat during expansion. 

According to the best authorities, the pressure varies in- 
versely as the T 9 (j power of the volume, if heat is neither added 
nor taken away by any outside source during the time the steam 
is being expanded in the steam engine cylinder. This is called 
adiabatic expansion of steam. 

The pressure varies inversely as the 17 /i6 power of the vol- 
ume, if the steam is kept dry at the temperature of saturation, 
during expansion, by means of a steam jacket outside the 
cylinder. 

When the pressure is considered to vary inversely as the 
volume it is called isothermal expansion. 

The isothermal curve is not exactly the correct curve to 
represent the expansion of steam, but it is the theoretical curve 
usually drawn on the indicator diagram, because it is so easy to 
handle and is also very nearly correct. 

The following formula gives the mean effective pressure 
according to isothermal expansion. 

M.E.P. = P 1 X Q+*****-' )-p t 

Absolute terminal pressure ^^X — - 



P\ = Absolute initial pressure. 
r = Ratio of expansion. 

P 2 = Absolute back pressure. 
M . E. P. = Mean effective pressure. 
Hyp. log. (hyperbolic logarithm), see page 

T,-.U1,-» "\T^ RR /-^J.rar. +V.Q tQvmi'nol o r-> A 4-V. c 



126. 



Table No. 66 gives the terminal and the mean effective 
pressure of steam expanded under any of these three different 
conditions. 



426 



NOTES ON STEAM. 



TABLE No. 66, — Constants for Calculating Mean 
Terminal Pressure of Expanding Steam. 



and 





ca 


V 


s 





u 






in 


w 




4> 




^3 


** 


** 


"o 


rt 


1/1 


4-* 


c 







u 





s. 


ri 


t- 1 .. 


fe 


u-^3 


£ 


©-S 


*** 4 


U C 




t> rt 


O 


J3 O. 




3W 


CJ 


£ 


% 


l 1 ^ 


%0 


l 3 /7 


% 


IX 


% 


1% 


6 /l0 


1% 


X 


2 


yio 


2y 2 


X 


2% 


x 


3 


3 /l0 


*% 


X 


4 


% 


5 


v« 


6 


¥t 


7 


^ 


8 


y 9 


9 


Yio 


10 


Y11 


11 


Y12 


12 


Vl3 


13 


VU 


14 


%5 


15 


yi6 


16 


%7 


17 


%8 


18 


%9 


19 


y 20 


20 



At Constant Tem- 
perature. 
(Isothermal Expan- 
sion). 



c 
6 



0.750 

0.700 

0.667 

0.625 

0.600 

0.500 

0.400 

0.375 

0.333 

0.300 

0.250 

0.200 

0.1667 

0.1428 

0.125 

0.1111 

0.1000 

0.0909 

0.0833 

0.0769 

0.0714 

0.0667 

0.0625 

0.0588 

0.0556 

0.0526 

0.0500 



3 ^ 

c 

CO 



u 



+ 



0.966 
0.950 
0.937 
0.919 
0.906 
0.846 
0.766 
0.743 
0.700 
0.662 
0.596 
0.522 
0.465 
0.421 
0.385 
0.355 
0.330 
0.309 
0.290 
0.274 
0.260 
0.247 
0.236 
0.225 
0.216 
0.208 
0.200 



Kept Dry at Tem- 
perature of Satura- 
tion. 
(Expansion in a 
Steam Jacketed Cy- 
linder) . 



,-. V 



a ~ U 



u 1-1 



0.737 

0.685 

0.650 

0.607 

0.581 

0.479 

0.378 

0.353 

0.311 

0.278 

0.229 

0.181 

0.149 

0.127 

0.110 

0.0968 

0.0865 

0.0782 

0.0713 

0.0655 

0.0605 

0.0563 

0.0526 

0.0493 

0.0463 

0.0438 

0.0415 



0.964 
0.947 
0.933 
0.914 
0.900 
0.839 
0.756 
0.732 
0.688 
0.648 
0.5S2 
0.506 
0.449 
0.405 
0.369 
0.340 
0.315 
0.293 
0.275 
0.259 
0.245 
0.232 
0.221 
0.211 
0.202 
0.193 
0.185 



Condensing by Work- 
ing in a Cylinder. 
(Adiabatic Expan- 
sion). 



Ph 1 



0.726 

0.673 

0.638 

0.593 

0.567 

0.463 

0.361 

0.336 

0.295 

0.262 

0.214 

0.167 

0.137 

0.115 

0.0992 

0.0862 

0.0774 

0.0696 

0.0631 

0.0578 

0.0533 

0.0494 

0.0459 

0.0429 

0.0403 

0.0379 

0.0359 



3 hU 



fc 



0.962 
0.944 
0.930 
0.911 
0.897 
0.833 
0.748 
0.723 
0.678 
0.637 
0.571 
0.495 
0.437 
0.393 
0.357 
0.327 
0.303 
0.282 
0.264 
0.248 
0.234 
0.222 
0.211 
0.201 
0.192 
0.184 
0.177 



NOTES ON STEAM. 427 

To Find the Mean Effective Pressure by the 
Preceding Table. 

Find the constant in the column corresponding to the con- 
ditions of expansion, and to the given cut-off. Multiply this by 
the absolute initial pressure, and the product is the average 
pressure. Subtract the back pressure and the remainder is the 
mean effective pressure. 

Example. 

Find the mean effective pressure for isothermal expansion 
when the engine is cutting off at one-quarter stroke. The initial 
pressure is 90 pounds absolute. The absolute back pressure is 
IS pounds. 
m Solution : 

~M. E. P. = 90 X 0.596 — 18 = 53.64 — 18 = 35.64 pounds. 
Note. — All such calculations must be made from absolute 
pressure (not gage pressure), and when determining the cut-off 
the clearance must be considered. 

Clearance. 

The clearance of an engine is usualty expressed as a per- 
centage of the piston displacement. The space between the 
piston and the cylinder head at the end of the stroke, also the 
cavities due to the steam ports, must be included in considering 
clearance. 

In high-class Corliss engines the clearance does not exceed 
2 % to 5 per cent., but in common slide-valve engines the clear- 
ance may go as high as 5 to 15 per cent. When clearance is 
taken into account the actual ratio of expansion is 



r 



R = Actual ratio of expansion. 
r = Nominal ratio of expansion. 

c = Clearance, expressed as a fractional part of the length 
of the stroke. 

Example. 

The nominal ratio of expansion is 4, and the clearance is 
5 per cent. What is the actual ratio of expansion? 

Note. — 5 per cent, is ^ioo — Y20 = 0.05 of the stroke. 
Solution : 
d 1-4- 0.05 

-J- + 0.05 

1 05 
R = ■ ' = 3.5 = Actual ratio of expansion. 
0.t> 



428 



SHOP NOTES. 



SHOP NOTES. 



Weight of a Grindstone. 

Multiply the constant 0.064 by the square of the diameter 
in inches and this product by the thickness in inches; the result 
is the weight of the grindstone in pounds. 

Example. 

Find the weight of a grindstone 30 inches in diameter and 
six inches thick. 

Solution : 

Weight = 0.064 X 30 X 30 X 6 = 346 pounds. 

Lathe Centers. 

In this country lathe centers are universally made 60 de- 
grees, but in Europe the most common practice is to make lathe 
centers 90 degrees. 

Morse Taper. 

The Morse Taper, which is so universally used for the 
shanks of drills and other tools, is given in 

TABLE No. 67.— Horse Taper. 



No. of 
Taper. 


Standard 
Plug Depth. 


Diameter of 

Plug at 
Large End. 


Diameter of 

Plug at 
Small End. 


Taper per 
Foot. 


1 

2 
3 
4 
5 
6 


2y% inch. 
9 9 a 

4. 1 " 

5ft * 
7% " 


0.475 inch. 
0.7 

0.938 " 
1.231 " 
1.748 " 
2.494 " 


0.369 inch. 
0.572 " 
0.778 " 
1.02 

1.475 " 
2.116 " 


0.600 inch. 
0.602 " 
0.602 " 
0.623 " 
0.630 " 
0.626 " 



For very complete information regarding the Morse Taper, 
see American Machinist, May 14, 1896. 

Jar no Taper. 

Inthe"Jarno Taper" the number of the taper gives the 
length of the standard plug in half-inches, and it gives the diame- 
ter of the small end in tenths of inches and the diameter of the 
large end in eighths of inches. For instance, a No. 8"Jarno 
Taper " is four inches long, one inch diameter at large end, and 
0.8 inch diameter at small end. The taper, of course, is 0.6 
inch per foot for all numbers. This is a very convenient sys- 
tem, and deserves adoption for its merits. The same taper is 



SHOP NOTES. 



429 



also very well adapted to the metric system, as 0.6 inch per 
foot is equal to 0.05 millimeter per millimeter. 

The following table is given to illustrate the system. The 
table could be extended to as large size tapers as are required 
for any work. 

TABLE No. 68.— Jarno Taper. 



Number of 
Taper. 


Length of 
Taper. 


Diameter of 

Large End of 

Taper. 


Diameter of 

Small End of 

Taper. 


1 
2 


1 


y 8 = 0.125 
% = 0.250 


A = 0.1 
1 =0.2 


3 
4 


IK 
2 


3/ s = 0.375 

y 2 = 0.500 


t 3 o = 0.3 

A = °- 4 


5 


2^ 


5/ 8 = 0.625 


^ =0.5 


6 


3 


34 = 0.750 


1 =0.6 


7 
8 


3^ 
4 


7/ 8 = 0.875 
1 = 1.000 


A = 0.7 

1 =0.8 


9 


4^ 


iy 8 = 1.125 


9 — 9 

TO w,i7 


10 


5 


1% = 1.250 


1 = 1.0 



■ This system of taper is described by "Jarno " in the Amer- 
ican Machinist, October 31, 1889. 

Marking Solution. 

Dissolve one ounce of sulphate of copper (blue vitriol) in 
four ounces of water and half a teaspoonful of nitric acid. 
When this solution is applied on bright steel or iron, the surface 
immediately turns copper color, and marks made by a sharp 
scratch-awl will be seen very distinctly. 

A Cheap Lubricant for Milling and Drilling. 

Dissolve separately in water 10 pounds of whale-oil soap and 
15 pounds of sal-soda. Mix this in 40 gallons of clean water. 
Add two gallons of best lard oil, stir thoroughly, and the solu- 
tion is ready for use. 

Soda Water for Drilling. 

Dissolve three-fourths to one pound of sal-soda in one pail 
full of water. 

Solder. 

Ordinary solder is an alloy consisting of two parts of tin 
and one part of lead, and melts at 360°. 

Solder consisting of two parts of lead and one part of tin 
melts at 475°. For tin work use resin for a flux. 



43° SHOP NOTES. 

Soldering Fluids. 

Add pieces of zinc to muriatic acid until the bubbles cease 
to rise, and the acid may be used for soldering with soft solder. 

Mix one pint of grain alcohol with two tablespoonfuls of 
chloride of zinc. Shake well. This solution does not rust the 
joint as acids are liable to do. 

When soldering lead use tallow or resin for a flux, and use a 
solder consisting of one part of tin and 1% parts of lead. 

Spelter. 

Hard spelter consists of one part of copper and one part of 
zinc. 

A softer spelter is made from two parts of copper and three 
parts of zinc. 

A spelter which will flow very easily at low heat consists of 
- % of Copper._4r>5£ of Zinc. an:. 9% :: Silver. When making 
any of these different kinds of spelter, melt the copper first in a 
black lead crucible and then put in the zinc after the copper 
has cooled enough to fumisl; just sunicient heat to melt the 
zinc, but not enough to burn it. Stir with an iron rod and 
after the metals have compounded and the compound is still 
molten, pour upon a basin of water. The metal in striking the 
water will form into small globules or shot and will so cooL 
leaving a coarse granular spelter ready for use. When pouring 
the metal let a helper keep stirring the water with an old broom. 

Alloy Which Expands in Cooling. 

Melt together nine pounds of lead, two pounds of antimony 
and one pound of bismuth. This alloy may be used in fastening 
foundation bolts for machinery into foundation stones. In such 
cases, collars or heads are left on the bolts and after the hole 
is drilled, in the stone a couple of short, small holes are drilled 
at an angle to the big hole : when the metal is poured in. it will 
flow around the bolts and also into these small holes, and it is 
almost impossible for the bolt to pull out. 

Caution. — When drilling holes in stone, water is always 
used, but this must be carefully dried out by the use of red-hot 
iron rods before the melted metal is poured in. If this pre- 
: ration is not taken the metal will blow out. making a poor job, 
and it may also cause accident by burning the hands and face 
of the man who is pouring it in. 

Shrinkage of Castings. 

General rule : 

% inch per foot for iron. 

3 i6 inch per foot for bras s . 

In small castings the molder generally raps the pattern 
more than the casting will shrink, therefore no shrinkage is al- 
lowed. Frequently castings are of such shape that the pressure 



SHOP NOTE-. 431 

of the fluid iron on some part of the mould is liable to make the 
sand yield a little and thereby cause the casting to be as large 
as, or even larger than the pattern. All such things a practical 
pattern maker takes into consideration when allowing for 
shrinkage in patterns. 

Case Hardening Wrought Iron and Soft Steel. 

Bone dust specially prepared for the purpose, or burnt 
leather scrap, is placed in a cast-iron box. together with the 
article to be hardened. Cover the top of the box with plenty of 
the hardening material in order to keep the air out. Heat the 
whole mass slowly in a furnace to a red heat from two to five 
hours in order that it may be uniformly and thoroughly heated 
through. A few iron rods about 5 i« inch in diameter are put in 
when packing the box. one end of the rod reaching about to the 
middle of the box. and the other end projecting out through the 
hardening material on top. When die box appears to have the 
right heat, these rods are pulled out one at a time, in order to 
judge of the heat in the center of the mass. When the box 
has been exposed to the fire the desired length of time, its 
contents are quickly dumped into cool water. 

Sieves of iron netting are laid on the bottom of the tub into 
which the case hardening material is dumped so that the hard- 
ened articles may be conveniendy taken up from the water by 
one of the sieves. The case hardening material itself is also 
taken out by another sieve which is of very fine netting and 
placed under the first one. The material is dried and used over 
again, and a little new material is added when repacking the 
boxes. 

When articles are well finished before hardening, this pro- 
cess gives a very fine color to both soft steel and wrought iron. 

Case hardening may also be effected by packing the articles 
in soot but this process does not give a nice color. 

Horn and hoof is also used for case hardening. Malleable 
iron may also be case hardened, but it requires careful handling 
in order to prevent its cracking and twisting out of shape. 



Case Hardening Boxes 

are made from cast-iron and are of various sizes. Small boxes 
may be made nine inches long, five inches wide, and four inches 
deep, and about one-fourth inch thick. They should be pro- 
vided with legs ?.t least one inch high so that the heat may get 
under the bottom as at the top. An ear having a rectangular 
hole through it should be cast under the bottom at each end of 
the box. This gives a chance to handle the box with a fork 
having flat prongs instead of taking it out of the hardening fur- 
nace with a pair of tongs, which is liable to break the box. 
as cast-iron is very inferior in strength when hot. 



432 SHOP NOTES. 

To Harden with Cyanide of Potassium. 

Heat the cyanide of potassium in a wrought iron pot until 
cherry red, and keep it so by a steady fire, immerse the pieces 
to be hardened from three to five minutes, according to their 
size and degree of hardness required, then plunge into cold 
water. Large pieces require more time than small ones, and 
the longer the article remains in the cyanide the deeper the 
hardening becomes. New cyanide gives the best color and 
cyanide previously used for hardening produces a harder surface. 



BLUE PRINTING. 
To Prepare Blue Print Paper. 

Dissolve two ounces of citrate of iron and ammonium in %% 
ounces of soft water. Keep this in a dark bottle. Also dissolve 
1% ounces of red prussiate of potash in 8X ounces of water and 
keep in another dark bottle. When about to use, mix (in a 
dark place) an equal quantity 7 of each solution in a cup and 
apply with a sponge or a camel's hair brush as evenly as pos- 
sible on one side of white rag paper (such as used for envelopes). 
Let it dry and put it away in a dark drawer. The paper must 
not be prepared in daylight but when taking prints it may be 
handled then, providing care is used to expose it as little as 
possible to the light before it is put into the printing frame. 

Blue Print Frame. 

Make a strong frame similar to a picture frame having a 
strong and thick glass. -Make a loose back, from boards about 
y 2 inch thick, which is held into the frame by four suitable 
catches so arranged that they press this back firmly and evenly 
against the glass. The surface next to the glass should be 
covered by three thicknesses of flannel in order to make a 
cushion so that the prepared paper and the tracing are kept close 
together when put in the frame. 

Blue Printing. 

The drawing must be made on transparent material, for 
instance, tracing cloth or tracing paper. Place the tracing in 
the frame with the side on which the drawing is made next to 
the glass. Place the prepared side of the sensitive paper against 
the back of the tracing. Put the loose back into the frame with 
the padded side against the prepared paper, and fasten it up so 
that both paper and tracing are kept firmly against the glass. 
Expose to sunlight from three to six minutes, according to the 
brightness of the sun. Take the sensitive paper out of the 
frame and quickly put it into a tub of clean cool water and wash 
it off, and the drawing will appear in white lines on blue 
ground. Hang the print up by one edge so that the water will 
run off, and let the print hang until dry. 



INDEX. 



A. 

Acceleration due to gravity, 276. 
Addition, 5. 

of decimals, 14. 

of vulgar fractions, 12. 

of logarithms, 75. 
Algebra, notes on, 63. 
Allowable deflection, 266. 
Alloy which expands in cooling, 

43°- 
Animal power, 318. 
Angular velocity, 299. 
Area of circles, 196. 

tables of, 209. 
Area of segments of circles, 199. 

of parallelograms, 196. 

of triangles, 193. 

of triangles, formulas for, 
176-177. 

of trapezoids, 196. 

of a circular lune, 202. 

of a zone, 202. 



Beams, deflection in, 254-266. 
transverse strength of, 233- 

254- 

to calculate size to carry a 
given load, 247. 

round wooden, 251. 

not loaded at the center of 
span, 252. 

loaded at several places, 253. 

placed in an inclined posi- 
tion, 254. 



Bearings, 369-374. 

area of, 369. 

allowable pressure in, 369. 

proportions of, 370-374. 
Belts, 326-338. 

arc of contact of, 332. 

angle, 337. 

cementing, 328. 

lacing, 327. 

length of, 329. 

horse-power transmitted by, 

3 2 9- 

crossed, 336. 

oiling of, 338. 

quarter turn, 336. 

slipping of, 237- 

tighteners on, 337. 

velocity of, 337. 
Bevel gears, 389. 

dimensions of tooth parts 
in, 391. 
Body projected at an angle up- 
ward, 279. 

in horizontal direction from 
an elevated place, 284. 
Blue print paper, how to prepare, 

43 2 - 
Blue printing, 432. 

C. 

Case hardening, 431. 
boxes, 431. 
by cyanide of potassium, 

43 2 - 
Cap screws, 407. 
Center of gravity, 292. 



433 



434 



INDEX. 



Center of gyration, 292. 

of oscillation, 292. 

of percussion, 292. 
Centrifugal force, 302. 
Chain Links, 323. 
Circles, 152. 

area and circumference of, 
209. 
Couplings, 369. 
Coupling bolts, 406. 
Compound proportions, 17. 
Cone, surface of, 204. 

volume of, 205. 
Constant for deflection, how to 

find. 264. 
Crane Hooks, 323. 
Cylinder, thickness of, 219. 

surface of, 203. 

volume of, 204. 
Cube root, 30. 

table of. 33. 
Crushing strength, 223. 
Clearance, 427. 

D. 

Debt paying by instalments, 89. 
Deflection in beams when loaded 
transversely, 234. 
allowable, 266. 
Derrick, proportions of a two ton, 

3 2 4- 
Difference between one square 
foot and one foot square, 

*93- 

Dimensions of U. S. standard 
screws, 405. 
of "VYhitword standard 
screws, 404. 
Diameter of tap-drill, 404. 

of screws at bottom of 
thread, 403. 
Discount or rebate, 83. 
Division. 6. 

of decimal fractions, 15. 
of vulgar fractions, 10. 
of logarithms, 78. 
Drawing, problems in geometri- 
cal, 184. 



E. 

Efficiency of machinery, 322. 
Ellipse, area of, 208. 

circumference of, 208. 
Elongation under tension, 215. 
Energy, kinetic, 287. 
Eye bolts, 407. 
Equations, 65. 

quadratic, 67. 
Equation of payments, 27. 

P. 

Factor of Safety, 274. 

Force energy and power, 285. 

of a blow how to calculate, 
28S. 

acceleration and motion 
formulas for, 288. 

centrifugal, 302. 
Formulas, 3. 
Fractions, addition of, 7, 14. 

subtraction of, 8, 14. 

multiplication of, 9, 15. 

division of, 10, 15. 

to reduce from one denom- 
ination to another, 11. 

to reduce a decimal to a 
vulgar, 13. 

to reduce a vulgar to a deci- 
mal, 12. 
Friction, 303. 

axle, 305. 

angle of, 306. 

co-efficient of, 304, 306. 

rolling, 304. 

rules for, 306. 

in machinery, 306. 

in pulley-blocks, 307. 
Frustum of a cone, 205. 

surface of, 205. 

volume of, 206. 
Frustum of a pyramid, 207. 

surface of, 207. 

volume of, 207. 
Fly wheels, 356. 

weight of, 356. 

centrifugal force in, 357. 



INDEX. 



435 



Fly Wheels, to calculate speed of 
bursted, 284. 

safe speed of, 360. 

safe diameter of, 360. 
French standard screws, 409. 

G. 

Gage for sheet iron and sheet 
steel, 145. 
for wire, 146. 
for twist drills and steel wire, 

148. 
for stubs, 148. 
Geometry., 149-152. 
Geometrical mean, 70. 
Gravity, 276. 

specific, 138. 
Gear teeth, 375-400. 

circular pitch of, 375. 
comparing circular and di- 
ametral pitch of, 380. 
cycloid form of, 382. 
dimensions of teeth, 380. 
diametral pitch of, 377. 
involute form of, 385. 
width of, 389. 
Gearing bevel, 389. 

how to calculate speed of, 

322. 
dimensions of tooth parts 

in, 380. 
dimensions of tooth parts 

in bevel, 391. 
worm, 395. 
dimensions of tooth parts 

in worm, 397. 
elliptical, 400. 
Grindstone, weight of, 140, 428. 
speed of, 321. 

H. 

Hauling a load, 318. 
Horse power, 317. 

of a steam engine, 317. 

compound or triple engine, 

3*7- 
of waterfalls, 318. 



Hot water heating, 422. 
Hydraulic, notes on, 413. 
Hyperbolic logarithms, 71, 126. 



I. 



I beam, strength of, 242. 
Impulse, 286. 

Interest, compound, computed 
annually, 22. 

semi-annually, 24. 

simple, 19. 

tables, 20-25. 

by logarithms, 81. 
Impulse, 286. 
Involute, 192. 
Inertia, 285. 

moment of, 293. 
Inclined plane, 308. 
International standard thread, 
410. 

J. 

Jarno taper, 428. 
table of, 429. 

K. 

Keys, proportions of, 368. 
Kinetic energy, 287. 

L. 

Lag screws, 408, 
Lathe centers, 428. 
Laws, Newton's, 276. 
Logarithms, 71. 

table of, 90. 

hyperbolic, 71, 126. 

table of, 126. 
Levers, 292. 
Lune, circular, 202. 
Lubricant for milling and drill- 
ing, 429. 

M. 

Machinery, efficiency of, 322. 
power required to drive, 319. 



-V- 



INDEX. 



Machinery, speed of. 320. 
Mathematics, notes on. 1. 
Manila ropes, transmission of 
power by, 344. 
transmission capacity of, 

345- 
preservation of, 347. 

weight of, 346. 

strength of, zzi 
Marking solution. \ : 3 
Mass. 2S6. 
Multiplication, 6. 

of decimals. 15. 

of vulgar fractions : 

::' ligaritiinis. "". 
Mechanics, 176. 
Mensuration, 193. 
Metric system of weights and 
measures, 1 3 5 

thread with inch divided 
lead-screw, how to cut, 
:: :: 
Modulus of elastic::; 213—216. 

how to calculate :::. 255, 

: : 5 
Moments, 292. 
Moment of inertia, 2 z. ] 

polar, 297. 
Momentum, 286. 
M: ;sr rarer. _:> 
Motion, Newton's laws of. 276 

down an inclined plane, 283 



1 i-ierian logarithms. 71, 126. 
Newton's laws of motion, z~z . 



Oiling of belts 13 3 



Parallelogram of forces, 316. 
ri"e:s:::. '.' 
Payments, equation of. 27. 
Pillars, safe load on 226—229. 
hollow cast iron. : : 3 



Pillars, wrought iron, 23 1. 

weight c: 228, 232. 
Polar moment of inertia, 297. 
Polygons. 149. 
Posts, wooden, 231. 
Power,, animal, 318. 

of man, 319. 

required to drive machinery, 

3 l 9- 
Pulley-blocks, 307. 

differential, 308. 
Pulleys. 348. 

how to calculate size of, 321. 

stepped, 350. 

correct diameter : : 15a 

for back-geared lathes. 3 -" _ 
Pyramids, frustum of a. 2 c " 

slantea area of a. 206. 

total area of a. 206. 

volume of a, 206. 
Pressure on bearings caused by 

the belt 333. 
I ressure of fluid in a vessel 413. 
Produce, weight of, 1 3 5 . 
Problems in geometrical draw- 
ing, 184. 
Progressions, 68. 
Proportion, 16. 

compound, 17. 



Quadratic equations, 67. 
Quantity of water discharged by 

pipes, 417. 
of water required to make 

any quantity of steam. 

422. 

R. 

Radius of gyration, 293. 
Radical quantities expressed 
without the radical sign, 

3 2 - 
Ratio, 16. 
Reciprocals, 32. 
table of, 33. 
Results of small savings, 26. 



INDEX. 



437 



Rope, manila, 344, 345, 347, 346, 
and 222. 
wire, 338, 340, 342, 343> and 
222. 



Safety, factor of, 274. 
Savings, results of small, 26. 
Sector of a circle, area of, 198. 
Segmemt of a circle, 199. 

length of arc of, 199. 

area of, 199, 201. 

of a sphere, volume of, 208. 
Sinking funds and savings, 84. 
Soda-water for drilling, 429. 
Solder, 429. 
Soldering fluid, 430. 
Subtraction, 5. 

of decimal fractions, 14. 

of vulgar fractions, 8. 

of logarithms, 76. 
Screws, 311, 401. 

"pitch," "inch pitch," of 
worms and, 401. 

French standard, 409. 

German standard, 410. 

international standard, 410, 
411. 

U. S. standard, 402. 

Whitworth standard, 404. 

diameter at bottom of thread 
of, 403. 

round and fillister head, 406. 

cap, 407. 
Screw-cutting by the engine 
lathe, 401. 

multiple threaded, 40 1. 
Shafting, 360. 

allowable deflection in, 362. 

classification of, 365. 

horse power of, 365, 366. 

not loaded at the middle be- 
tween bearings, 361. 

torsional strength of, 363. 

torsional deflection of, 364. 

transverse strength of, 360. 

transverse deflection of, 361. 
Shafts for idlers, 367. 



Shearing strength, 272. 
Shop notes, 428-432. 
Shrinkage of castings, 430. 
Spelter, 430. 
Speed of machinery, 320. 

of handsaws, 320. 

of drilling-machines (ircjn), 
320. 

of emery-wheels and straps, 
321. 

of grindstones, 321. 

of lathes (for iron and wood), 
320. 

of milling machines, 320. 

of planers, 320. 
Specific gravity, 138. 
Sphere, surface of a, 207. 

volume of a, 208. 
Steam, notes on, 418-427. 

properties of saturated, 419 

expansion of, 425-426. 

weight of, required to boil 
water, 424. 

heating, 421. 
Strength of materials, 213-275. 

tensile strength, 213. 

crushing, 223. 

transverse, 233. 

torsional, 266. 

shearing, 272. 

cast iron, 214. 

wrought iron, 214. 

cylinders, 219. 

flat cylinder heads, 220. 

of dished cylinder heads, 
220. 

of bolts, 218. 

of hollow sphere, 221. 

of chains, 222. 

of wire rope, 222. 

of manila rope, 222. 

of beams when section is 
not uniform, 243. 

of square and rectangular 
wooden beams, 245. 
Square root, 29. 

table of, 33. 

by logarithms, 78. 



43* 



INDEX. 



Tank, to calculate number of 

gallons in a, 202. 
Tensile strength, 216. 
Torsional deflection. 271. 

in hollow round shafts, 271. 
strength, 266. 

strength in hollow round 
shafts, 270. 
Transverse strength, 233. 

U. 

Upward motion. 279. 

U. S. Standard screws, 402. 

V. 

Value of various metals, 139. 
of the trigonometrical func- 
tions for some of the most 
common angles, 156. 
low pressure steam for heat- 
ing purposes. 422. 
Velocity, angular, 299. 
of efflux, 413-414. 
of water in pipes, 415. 

W. 

Water, measure of, 13S. 
weight of, 13S. 



Water required to condense one 

pound of steam. 423. 
Weights and measures, 133. 

metric system of. 135. 
Weight of iron, square or round, 

143- 

flat. 144. 

of any shape of section, 141. 

of sheet-iron of any thick- 
ness. 141. 

of steel, 142. 

of casting from weight of pat- 
tern, how to calculate, 141. 

metals not given in the 
tables. 141. 

of cast-iron balls, 142. 

and value of metals. 139. 

of various materials. 140. 

of liquids. 140. 
Wire gages. 146. 
Wire rope transmissions, 338. 

capacity of, 340. 

deflection in, 342. 

weight of. 343. 

strength of, 343. 



Z. 

Zone circular, 202. 



CLASSIFIED CATALOGUE OF 

BOOKS ON STEAM, STEAM ENGINES, Etc 

FOR SALE BY 

D. VAN NOSTRAND COMPANY, 

23 Murray and 27 Warren Sts., New York. 



BOILERS. 

Barr. Practical Treatise on High Pressure Steam Boilers, including 
Results of Recent Experimental Tests of Boiler Material, etc. 8vo. Illus- 
trated. Indianapolis, 1893. $3.00 

Barrus. Boiler Tests : Embracing the results of one hundred and thirty- 
seven evaporative tests, made on seventy-one boilers, conducted by the 
author. 8vo. Boston, 1895. $5-oo 

Christie. Chimney Design and Theory. A book for Engineers and Archi- 
tects, containing all data relative to Chimney Designing. Illustrated 
with numerous diagrams and half-tone cuts of many famous chim- 
neys. 8vo, cloth. Illustrated. New York, 1899. $3.00 

Courtney. The Boiler Maker's Assistant in Drawing, Templating, and 
Calculating Boiler Work and Tank Work, with rules for the Evapora- 
tive Power and the Horse Power of Steam Boilers, and the Proportions 
of Safety Valves, and Useful Tables of Rivet Joints of Circles, Weights 
of Metals, etc. Revised and edited by D. K. Clark, C.E. Illustrated. 
London, 1898. (Weale's Series.) $0.80 

The Boiler Maker's Ready Reckoner. With examples of Practical 
Geometry and Templating, for the Use of Platers' Smiths, and Riveters. 
Revised and edited by D. K. Clark. 3d edition. London, 1890. (Weale's 
Series.) $1.60 

Davis. A Treatise on Steam-Boiler Incrustation, and Methods for Pre- 
venting Corrosion and the Formation of Scale ; also a Complete List 
of all American Patents issued by the Government of the United States 
from 1790 to July 1, 1884, for Compounds and Mechanical Devices for 
Purifying Water, and for Preventing the Incrustation of Steam Boilers. 
65 engravings. 8vo. Philadelphia, 1884. $2.00 

Foley, Nelson. The Mechanical Engineer's Reference Book for Ma- 
chine and Boiler Construction, in two parts. Part I., General En- 
gineering Data. Part II., Boiler Construction. With 51 Plates and 
numerous illustrations specially drawn for this work. Folio, half mor. 
London, 1895. $25.00 

Horner. Plating and Boiler Making. A Practical Handbook for Work- 
shop Operation, including an Appendix of tables by A Foreman Pattern 
Maker. 338 illustrations. i2mo. London, 1895. $3.00 



LIS'i 01 BOO. 

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Traill. Boilers: Their . - T . A Hi:. : 

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tioas |d e i .-..:;■. : : tnor. London, 1S96. I ■ - - 

Triplex. Marine Boilers. A. Treatise the lasses and Prevention of 

7 or Priming Rean beat Mnnnm im m fflastri 

i2mo. Sunderland, 1S99. fa - 

Watson. Small Engines and Boilers. A AAr.ua. A A:;.:: ar_: 
Specific Diiectioos for the Coastiact] m A Saiall Steam Eagiaes aad 
Boilers of Modern Types A : m Eve a : rs t: : t : lom : : modd s : e 5 
i2mo, cl::a 1 Aa5:rated with numer: _s d.ai-rarr.s sad hatf-tune cats 
Mew York. 1899. >: :: 

The intention of the author in — r.:.r. r this weak has :eer. :: Ear- 
nish specific liiecdoas and correct dimer.E. : :e: plans for small 
engines and : :ilers 

Wilson. A Treatise on Steam Boilers: TLei: ; enrrr A: : : -.::- 
and Econonnii. :: .r Enlarged aad :!As:rated from tie 7_i Zz 
fish edhioa by J 7 Fbthei ezbh New York, 1897. 5: -• : 

Boiler and Factory Chimneys; Their T'raarr.t I:— er and 5:a":Arv 

A e :..■ . : r : zai : Loadoa, 1892 j : : 

FUELS. 

Abbott. Treatise on Fuel. Founded :>n the trigina] Treatise A A 
rumens. lAastrared. i6mc. New York 1891. 5: ;; 

Barr. Practical Treatise on the Combustion of Doal : 1 desmr;- 

5 A varioas n:e: : : ie :e5 ::; the r_::: :~_: Generation of 
Hea: by the Combusn::: A Fad whether Sofid, liquid, or Gaseoas. 
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Clark and Williams. Fuel: As rombusrAr. arA Z :- sis:: . : 

■ . t5 ::' Treatise 00 the lAnAustion of Coal and the Irnomy 
of Fuel. With extensive :::.::> hi recent practice . the A rust::: 
and Economy of Fad Ai. 1 : ke Wood. Peal I e troJeam, etc 4th 
edition. i:mo. Londc 5: 90 

Bbdgetts. Liquid Fuel fan Methanical and Industrial Purposes. IHns- 
trated v : London. 1 So: 5: 5 : 

?:A.lips. Fuels: Solid, Liquid, a ua-~ 

': the se oJ Er. Ar.eers : " 

£o.So 



LIST OF BOOKS. 

Sexton, A. H. Fuels and Refractory Materials. 8vo. Cloth. London, 
1897. $2.00 

Williams. Fuel : Its Combustion and Economy. Consisting of an 
Abridgment of " A Treatise on the Combustion of Coal and the Pre- 

. vention of Smoke." With extensive additions by D. Kinnear Clark. 
4th edition. London, 1891. $1.50 

GAS ENGINES. 

Clerk. The Theory of the Gas Engine. 2d edition, with Additional 
Matter edited by F. E. Idell. i6mo. New York, 1891. $0.50 

The Gas Engine. History and Practical Working. With 100 illus- 
trations. 6th edition. i2mo. New York, 1896. $4.00 

Donkin. A Text-Book on Gas, Oil, and Air Engines : or Internal Com- 
bustion Motors without Boiler. 154 illustrations. 8vo. London, 1896. 

Goodeve. On Gas Engines : with Appendix describing a Recent Engine 
with Tube Igniter. i2tno. London, 1887. $1.00 

Hiscox, Gardner D. Gas, Gasoline and Oil Vapor Engines. A New 
Book Descriptive of Their Theory and Power. Second edition, revised 
and enlarged. 8vo, cloth. Illustrated. New York, 1898. $2.50 

HEAT. — THERMODYNAMICS. 

Anderson. On the Conversion of Heat into Work. A Practical Hand- 
book on Heat Engines. 3d edition. Illustrated. i2mo. London, 
1893. #2.25 

Box. Treatise on Heat as Applied to the Useful Arts, for the use of 
Engineers, Architects, etc. 8th edition. i2mo. London, 1895. $5-°° 

Larden. A School Course on Heat. Illus. i2mo. London, 1894. $2.00 

McCulloch. Elementary Treatise on the Mechanical Theory of Heat and 
its application to Air and Steam Engine. 8vo. New York, 1876. $3.50 

Maxwell. Theory of Heat. New edition, with Corrections and Addi- 
tions by Lord Rayleigh, Sec. R. S. Illustrated. i2mo. New York, 
1897. $1.50 

Peabody. Thermodynamics of the Steam Engine and other Heat En- 
gines. 8vo. New York, 1898. $5-oo 

Rontgen. The Principles of Thermodynamics. With special Applica- 
tions to Hot Air, Gas, and Steam Engines. With additions from Profes- 
sors Verdet, Zeuner, and Pernolet. Translated, newly and thoroughly 
revised and enlarged by Professor A. Jay Du Bois. 732 pages. 3d edi- 
tion. 8vo. New York, 1896. $5.00 



D. VAN NO STRAND COMPANY. 

Tyndall. Heat considered as a mode of Motion. 6th edition. i2mo. 

New York, 1890. $2.50 

Williams. On Heat and Steam : embracing New Views of Evaporization, 

Condensation, and Expansion. Illus. Svo. Philadelphia, 18S2. $2.50 
Wood. Thermodynamics, Heat Motors, and Refrigerating Machines. 

Revised and enlarged edition. Svo. New York, 1S95. $4.00 

HOISTING MACHINERY. 

Colyer. Hydraulic, Steam and Hand Power-Lifting and Pressing Ma- 
chinery. 72 large plates. Svo. London, 1892. $10.00 

Glynn. Treatise on the Construction of Cranes and other Hoisting Ma- 
chinery. 7th edition. Illustrated. London, 1SS7. $0.60 

Marks. Notes on the Construction of Cranes and Lifting Machinery. 
i2mo. London. 1S92. $1.00 

Towne. A Treatise on Cranes, descriptive particularly of those designed 
and built by the Yale and Towne Manufacturing Company, owning and 
operating the Western Crane Company , including also a description of 
light hoisting machinery as built by the same makers. Svo. New York, 
1S83. $ 1 - 00 

Weisbach and Hermann. The Mechanics of Hoisting Machinery, in- 
cluding Accumulators, Excavators, and Pile-drivers. A Text-book for 
Technical Schools and a guide for Practical Engineers. Authorized trans- 
lation from the second German edition by Karl P. Dahlstrom. 177 illus- 
trations. Svo. New York, 1893. $3-75 

ICE-MAKING MACHINES. 

Dixon. Manual of Ice-Making and Refrigerating Machines. A Treatise 
on the Theory and Practice of Cold-Production by Mechanical Means. 
i6mo. St Louis, 1894. $i.oc 

Leask. Refrigerating Machinery. Its Principles and Management. 
With numerous illustrations. Svo. London, 1S94. $2.00 

Ledoux. Ice-Making Machines : the Theory of the Action of the Various 
Forms of Cold-producing or so-called Ice-Machines. Translated from 
the French. 24S pages and numerous tables. i6mo. New York, 1892. 

#0.50 

Redwood. Theoretical and Practical Ammonia Refrigeration. A Prac- 
tical handbook for the use of those in charge of refrigerating plants. 
Illustrated with numerous Tables. i2mo. New York, 1S96. $1.00 

Wallis-Tayler. Refrigerating and Ice-Making Machinery. i2mo, cloth. 
Illustrated. London, 1896. $3-Qo 



LIST OF BOOKS. 



INDICATORS. 

Bacon. Treatise on the Richards Steam Engine Indicator. With a 
Supplement, describing the latest Improvements in the Instruments for 
Taking. Measuring, and Computing Diagrams. Also an Appendix, con- 
taining Useful Formulas and Rules for Engineers. 23 diagrams. 4th 
edition. i6mo, flex. Xew York. 1SS5. Si. 00 

Ellison. Practical Applications of the Indicator. With reference to the 
Adjustment of Valve Gear on all Styles of Engines. 2d edition. Svo. 
100 engravings. Chicago, 1897. S2.cc 

Hemenway Indicator Practice and Steam Engine Economy. With 
Plain Directions for Attaching the Indicator, Taking Diagrams, Comput- 
ing the Horse-Power, Drawing the Theoretical Curve. Calculating Steam 
Consumption. Determining Economy, Locating Derangement of Valves, 
and making all desired deductions ; also, Tables required in making the 
necessary computations, and an Outline of Current Practice in Testing 
Steam Engines and Boilers. 6th edition. i:mo. Ne-.v York. iS:S. 

S2.CC 

Le Van. The Steam Engine Indicator and its Use. A Guide to Practi- 
cal Working Engineers for greater economy, and the better Working of 
Steam Engines. iSmo, boards. New York. 1S96. So. 50 

The Steam Engine and the Indicator : Their Origin and Progressive 

Development, including the most recent examples of Steam and Gas 
Motors, together with the Indicator, its Principles, its Utility, and its Ap- 
plication. Illustrated by 205 engravings, chiefly of Indicator-cards. Svo. 
Philadelphia, 1S90. £4.00 

Porter. A Treatise on the Richards Steam Engine Indicator, and the 
Development and Application of Force in the Steam Engine. 5th edi- 
tion, revised and enlarged. Svo. London, 1S94. Sj.c: 

Pray. Twenty Years with the Indicator. Being a Practical Text-book 
for the Engineer or the Student, with no Complex Formulae. With 
many illustrations and rules as to the best way to run any Steam Engine 
to get the most economical results. How to Adjust Valves and Valve 
Motions Correctly. Full directions for working out Horse-Power, the 
Amount of Steam or Water per Horse-Power, Economy and Fuel. Ex- 
tended directions for Attaching the Indicator, what Motions to use and 
those not to use. Full directions for Computation of Power by Planim- 
e:er and other methods, with many tables and hints. Svo. New York, 

1896. £ 2 -5° 



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